# Homework Help: Finding the range of a projectile at an angle

1. Oct 29, 2012

### Precepts

1. The problem statement, all variables and given/known data

Calculate the range if a sniper stands and raises his barrel to an angle of 30 degrees above the horizontal with the end of the barrel 2.0 m above the ground. The muzzle velocity of his rifle is 800m/s.

Known variables:

U = 800ms^-1
Uv = 400m/s
Uh = 692.8m/s
a = -9.8m/s

2. Relevant equations

V = u + at
v^2 = u^2 + 2as
s = ut + 1/2at^2

3. The attempt at a solution

Sh = Uh x t

In order to find t, I doubled the time to the apogee:

Vv = u + at
0 = 400 x -9.8t
9.8t = 400
t = 40.8 sec. (apogee)

Therefore time of flight = 81 sec. (This is obviously wrong... using this in the range equation gives too large a number). So I don't know where to go from here. The launching at an angle + a height is confusing me, I'm just not seeing it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 29, 2012

### PeterO

Just where did you want to know the range?
To a point also 2m off the ground [ that's what you have been working towards ] or to a point on the ground, ie 2m below the launch height. [given the speed involved this has to be about 2√3 metres further than what you were going to calculate.

3. Oct 29, 2012

### Precepts

Right, I'm not sure where the 2m comes into it. I want to know the horizontal range for when it hits the ground. I'm not sure how to go about doing this. Thanks.

4. Oct 29, 2012

### PeterO

When you chose to (inaccurately) double the "time to apogee", you were using the principle that a projectile takes just as long to go up as it takes to come down. That is true only if the projectile comes down to the same height as it left from.
The barrel of the gun was 2m above the ground - the bullet comes all the way back to ground level - so you said.

By the way 40.8 x 2 is not equal to 81

5. Oct 29, 2012

### Precepts

Yes, I understand this. Can you calculate to include the 2m?

6. Oct 29, 2012

### Precepts

Or at least show me how...?

7. Oct 29, 2012

### PeterO

By "can you calculate to include the 2m" were you asking whether it could be done or were you asking whether I could calculate it?

To the latter - yes I can.

To the former - yes it can be.

Can you calculate the time taken for a ball to reach the ground when thrown up from the roof of a 20m building? (that is a general question, but if you want some practice, lets say it was thrown at 20m/s at an angle of 30o with the horizontal. To make the calculations really easy, take g = 10)

8. Oct 29, 2012

### PeterO

To allow for the 2m height, the final vertical displacement of the bullet is 2m down.

9. Oct 29, 2012

### Precepts

Thanks, could you work the solution for me? I just want to see how you did it.

Edit - Nvm, I solved it.

Last edited: Oct 29, 2012
10. Oct 29, 2012

### PeterO

I will use g = 10 for simplicity. You can do it "properly" using g = 9.8

Initial velocity 800 m/s means initial vertical is 400 m/s up.

line up the variables - I use v,u,a,s,t

take up as positive:

u = 400
v = not wanted
a = -10
s = -2
t = ?

I want the motion equation without v ..... s = ut + 0.5 at2

-2 = 400t - 5t2

so 5t2 - 400t - 2 = 0

use general solution to quadratic equation

t = [400 ±√(160000 + 40)]/10

t = [400 ±√160040]/10

t = [400 ± 400.05]/10

t can't be negative so

t = 800.05/10 = 80.005

range is thus Horizontal speed times 80.005 ie 400√3 x 80.005 = whatever

11. Oct 29, 2012

### Precepts

Thanks for that, I didn't use the quadratic equation and got the correct answer. I forgot about it.

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