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# how do you find height of a ball when you know velocity of ball

Using the given equations and values, find the height H of the ball at position 2.  TE!=TE2       v1=0m/s  h1=15m v2= 15m/s     h2= ?    TE=KE + UE KE = 1/2*m*v^2 UE = m*g*h g = 9.8m/s^2 Please help!!!

## Research, Knowledge and Information :

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## Suggested Questions And Answer :

### Calculate the hieght

Gravity (g=9.81m/s/s approx) causes the ball to follow a parabolic trajectory (assuming negligible air resistance) and s=u(y)t-½gt^2, where t=1.8s is the time in the air, u(y) is the vertical component of the ball's velocity and s is the vertical distance at the final destination of the ball, so s=h (where h is the height of the ball when it's caught). The horizontal distance=u(x)t where u(x) is the horizontal component of the ball's velocity. So 43.3=1.8u(x) and u(x)=43.3/1.8=24.06m/s approx.  The balls's velocity is √((u(y))^2+(u(x))^2)=26. We can find u(y)=√676-24.06^2=9.87m/s approx. We can now find out how long it would have taken for the ball to fall to the ground if it hadn't been caught. The final height of the ball would be zero, so 0=u(y)t-½gt^2, and t=0 or 2u(y)/g=2*9.87/9.81=2.011. The ball reaches its maximum height at half this time=1.01s approx. The actual height is 9.87*1.01-4.905*1.02=4.95m approx. The height of the ball when caught was about 1.87m=9.87*1.8-½g*3.24. The question is slightly ambiguous: it could mean the maximum height (4.95m) of the ball before it's caught or the height it was caught (1.87m).

### how do you find height of a ball when you know velocity of ball

???????????? velosity ?????????? it be SPEED...mite be vertikal speed, depend...yu sae "hite" that impli UP

### If a ball is thrown straight upwards with a velocity of 80 ft/s...

s = 80t - 16t^2 One way to do it: The ball starts at a height of 0 and ends at a height of 0.  Find when that happens. 0 = 80t - 16t^2 0 = t(80 - 16t) 80 - 16t = 0 80 = 16t t = 5 The ball will return to the ground at t = 5 seconds.  The maximum height will happen halfway between the start at t = 0 and the end at t = 5. Answer:  2.5 seconds. . Another way: s = 80t - 16t^2 is an upside-down parabola.  The maximum height happens at the top of the parabola, where the parabola is flat (slope = 0, first derivative = 0). s ' = 80 - 32t 0 = 80 - 32t 32t = 80 t = 2.5 Answer:  2.5 seconds.

### Find the earliest interception and latest interception.

The mathematical equation for the warhead's height after time t seconds is h=490000-4107t. At t=0, h=490000. The intercepting missiles are delayed by 10 seconds so the equation for their height is h=6115(t-10) so that when t=10, h=0. The warheads and interceptors are assumed to moving at uniform velocity, so graphically the height equations for both missiles are straight lines. Where they cross represents interception, which must take place before the warheads explode at h=16000. t=(490000-16000)/4107=115.41secs. To find out the earliest time for interception we have 6115(t-10)=490000-4107t, when the missile and interceptor are at the same height. 6115t-61150=490000-4107t; 10222t=551150, t=53.92 secs. (h=268,550.56 ft approx.) The latest time for interception is t=115.41 secs when the missiles explode. To reach a height of 16000 feet it takes the interceptor 16000/6115=2.62 seconds. Add on the 10 seconds to launch=12.62 secs. It takes the missile 115.41 secs to reach 16000 feet so the interceptor must be launched no later than t=115.41-12.62=102.79 seconds. To illustrate this using Excel we could draw the two graphs initially, h being the vertical axis and t the horizontal: h=490000-4107t and h=6115(t-10). This represents the earliest time of interception. Then we need to "slide" h=6115(t-10) to the right to cross the other graph where h=16000. Where this shifted graph intercepts the t axis should correspond to 102.79 seconds. Some scaling down will be necessary for h perhaps measuring height in 1000's or 10000's of feet. The t axis should be scaled to accommodate about 120 seconds and of course the origin is on the extreme left. If a graph is not used, Excel formula should be used to create a table of h and t values suitably spaced. =IF(A1<10,0,6115*(A1-10)) can be used as a formula for the interceptor. =IF(490000-4107*A1<16000,16000,490000-4107*A1) can be used as a formula for the missile. The graph can be inserted using the tabulated data. Cell A1 contains t, so column A contains all the values of t you want to use. B1 and C1 correspondingly contain data for h for the two lines.

### How do you calculate volume of a cylinder with ends of different diameter?

A cylinder with different end diameters is a conical cylinder. This means you find the volume of the cylinder by subtracting the volume of a cone from the volume of a larger cone where the smaller cone forms the top section of the larger cone. If you know the two diameters, you can work out the height of the large cone by treating the cones like similar isosceles triangles. The large cone is the triangle ABC, with apex A, and the smaller cone is the similar triangle ADE, where D Is on AB and E on AC. The side DE is parallel to BC. If we call DE and BC the bases of the two triangles, then angles ADE, ABC, AED and ACB are all equal. Drop a perpendicular from A to BC. That's the height of the larger cone. The ratio of the bases is the same as the ratio of the heights, and the two bases represent the two diameters of the ends of the conical cylinder, represented by the trapezium DECB.  If h is the height of the conical cylinder, and the diameter of the small end is d1 and that of the bigger end is d2, it can be shown by similar figures that h1, the height of the smaller cone,=hd1/(d2-d1). The height of the larger cone is h1+h. The volume of the large cone is (pi)(d2/2)^2(h1+h)/3 and the volume of the smaller one is (pi)(d1/2)^2h1/3. The difference is the volume of the conical cylinder. I'll leave you to do the substitutions!

### a kicker punts a football 3 feet above the ground with an initial velocity of 47 feet per second.

ft/sek is SPEED tu get hite av ball, need the angel from horizontal the ball start at ie, need vertikal komponent av VELOSITY if horizontal, get time tu hit ground from hite=(1/2)gt^2 =3 ft time=root(2*3/32.2)=root(6/32.2)=root(0.186)=0.4317 sekond

### Calculate how high above the ground level is the ball caught

The ball can be struck at any angle, but it will travel furthest horizontally if the angle is 45 degrees. In the absence of any further information, let's assume the batsman hits the ball at an angle of 45 degrees. The ball's velocity can be resolved into two components: vertical and horizontal. If u is the speed imparted to the ball by the bat, the vertical and horizontal components are each u√2/2 (or u/√2). The equations for motion under gravity only apply to the vertical component and the horizontal component is just u√2/2. The equation for vertical distance is s=ut√2/2-gt^2/2 where g is the acceleration of gravity (9.81 m/s/s). The equation for horizontal distance is s=ut√2/2, because there is no horizontal component for gravity. The fielder (catcher) intercepts the ball a height h above the ground, so h=ut√2/2-gt^2/2 where t=1.9 seconds. So h=1.9√2u/2-4.905*3.61=1.3435u-17.707 approx. This equation relates h and u. The horizontal distance travelled when the ball is caught is 43.7=1.9√2u/2, from which u=32.5269 m/s. So h=25.99 m which seems excessive for a fielder playing on a level playing field. We assumed the angle the bat struck the ball was 45 degrees. Now let's say the angle was ø degrees to the level ground. The horizontal component of u is ucosø and the vertical is usinø. If ø is 90 degrees, for example, there is no horizontal component and the horizontal distance travelled by the ball is zero. Rewrite the equations: h=utsinø-gt^2/2, and s=utcosø, so u=s/(tcosø) (ø≠90 degrees) and h=stanø-gt^2/2. (When ø=45 this comes to 43.7-17.707=25.99 m as we saw earlier.) Therefore h=43.7tanø-17.707 metres. Note that h only depends on ø so it's independent of the speed of the ball. From this tanø=(17.707+h)/43.7. If 0≤h<4m (just about possible for a tall fielder), tanø<0.4967, and 22<ø<26.4. (The vertical component of the velocity is usinø. If the ball is allowed to fall to the ground without being caught, h=0=utsinø-gt^2/2 so t(usinø-gt/2)=0 from which t=0 (when the batsman hits the ball) and usinø=gt/2=4.905*1.9=9.32 approx. The horizontal component of the velocity is ucosø and the ball will travel a distance s=utcosø. Interestingly, if we put s=43.7 and t=1.9, ucosø=23 exactly. Do we therefore assume that the fielder was standing at the actual spot where the ball would have landed if it hadn't been caught? If so, the ball was caught at height zero. We can also calculate ø because usinø/ucosø=tanø=9.32/23=0.4052 approx and ø=22.06 degrees approximately, and u=23secø=24.82 m/s approx.) The red curve (almost a straight line) shows the height in metres against the angle of projection ø. The blue line is the "human cut-off point", the upper limit, for catching the ball after 1.9 seconds at a distance of 43.7m from the batsman. It also shows the height when ø=45º.

### merry go round.10 ft radius.8 oz steel ball resting @ perimeter.Groove in floor to center. MGR spinning 1 rev/sec. How much impulse force to push ball inside groove to 0 velocity as it passes center?

First, we need to be clear about what forces are acting when an object is in circular motion. In this case, there are no vertical forces other than the force of gravity keeping the roundabout, or merry-go-round, and the ball on the ground. The only horizontal force, neglecting any form of friction, is the centripetal force pulling the ball to the centre. This force depends on the speed of rotation and is mv^2/r, where v is the linear tangential speed of the ball or mw^2r where w is the angular speed, and r is the radius and m is the mass of the ball. The angular speed is 360 degrees per second (1 revolution per second) or 2(pi) radians per second. The actual speed of the ball is 2(pi)r feet per second, about 62.83ft/s or 42.84 mph. The speed is constant but the velocity varies because it's changing direction as a result of the centripetal force. Since s, the sector arc, is r times the sector angle ø, the tangential speed is rw. So rw=v and v^2=r^2w^2 and v^2/r=rw^2. Hence we can express the force in two ways, and force is mass times acceleration. Now, this is where we have to be careful. The only force is the centripetal force, so what is centrifugal force? After all, we've all experienced an outward push whenever we move in a circle. Surely the ball experiences such an outward force? Well, this is an illusion, because there's only one force. The apparent centrifugal force is only an equal and opposite reaction to the centripetal force. It's sometimes called reactive centrifugal force. It's actually the inertial resistance of an object to being constrained to move in a circle. The object really wants to stay at rest or continue moving at the same velocity (that is, same speed and direction), but if the object is constrained from doing so by, say, the walls of the rotating body, the inertial resistance appears to be a force pressing the object against the walls. In a vehicle, for example, executing a turn, a passenger may fell pushed against the side of the vehicle, while what's really being experienced is that the passenger is trying to move straight ahead, but can't do so because of constraints imposed by the side of the vehicle. So, since the ball is not constrained by a wall, it will naturally be flung out tangentially as the merry-go-round starts to rotate. If a force is applied it must be equal to mw^2r=0.5*4(pi)^2*10=197.4ft-lbs and this force has to be continuously applied as the machine rotates, otherwise the ball will simply be ejected. The alternative is a barrier or wall that moves with the merry-go-round. The groove will not affect this, because the groove is rotating too and will move "out of the way" as the ball moves tangentially unless restrained by the continuously applied force. There's no radial velocity so applying a radial impulse will just give the ball some brief change of momentum towards the centre, but as the system is rotating it will appear to an observer watching the merry-go-round that the ball is curving in towards the centre, the so-called Coriolis effect. The rotational speed of the ball will change because it has less distance to travel in one revolution. Think about stopping the system and giving the ball a brief push towards the centre. It will continue in a straight line along its groove directly to the centre having achieved some constant velocity as a result of the impulse. Now consider the system in motion. From the point of view of an observer riding with the ball, it will appear to move at constant speed to the middle. The centripetal force is taking care of the rotational velocity independent of this radial motion, So, in my view, we have two independent things happening here and an impulse is always going to increase the ball's speed to a constant, so that it never has a zero value when it reaches the centre. I hope you appreciate that the above explanation is my understanding of the mechanisms at work, and I don't speak as an authority on the subject!

### If the volume of a container is 184 cm3, & the height is 8cm, what is the container's base area?

to find the base area you need to divide the volume by the height (184:8=23), but there is no way you can find the length and the width(and you don't need, you've found the base area already)

### how many 1/2" candies will fit in a 41/2" x 4" bowl?

Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.  The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes. Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx. How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement. At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container. Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.
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