# Kinematics Problem

1. Jul 10, 2015

### FredericChopin

1. The problem statement, all variables and given/known data
http://imgur.com/BPPbwpu

2. Relevant equations
s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation.

3. The attempt at a solution
Let u be the velocity at which the projectile was launched.

It is given that sx = symax. So:

sx = symax = u*cos(θ)*t

Thus:

symax/t = u*cos(θ)

Now, in the y-direction, the projectile was given initial velocity u*sin(θ), and at the top of its trajectory (where it has maximum height symax), it has final velocity 0 m/s. Thus:

symax = ((0 + u*sin(θ))*t)/2

, which means:

symax/t = u*sin(θ)/2

However, since symax/t = u*cos(θ) :

u*cos(θ) = u*sin(θ)/2

Thus:

cos(θ) = sin(θ)/2

2*cos(θ) = sin(θ)

2 = sin(θ)/cos(θ)

2 = tan(θ)

θ = tan-1(2) = 63.4349°

However, this was marked wrong.

Can somebody see what could have gone wrong?

Thank you.

2. Jul 10, 2015

### ehild

The time to reach the highest point is not equal to the whole time of flight.

3. Jul 10, 2015

### FredericChopin

Ahh... I got it. It takes only half the total travel time to reach the top of the trajectory. Thank you.
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