# Projectile motion - Kinematics

## Homework Statement

At 75% its maximum height, the speed of a projectile is four-fifths its initial speed. What is the angle of the initial velocity vector with respect to the vertical?

I really have no idea how to do this. Can you give me the idea on how to start or how to do it. (just please don't tell me the answer)

Thanks a lot!

## Answers and Replies

Shooting Star
Homework Helper
At the max height, there is no component of velo in y dircn. vx remains constant. vy is given by the eqn vy^2 = vy_0^2 - 2gh.

Do you understand something now?

if i will use that equation there will be 2 unknown variables. the initial velocity of y and the height..

how can i use the given 75% its max height and the 4/5 speed of the projectile

As shooting star wrote,

$${v_{y0}}^2 = 2gy_\textrm{max}$$

Try working with that equation and

$${v_{y2}}^2 = {v_{y0}}^2 - 2gy_{2}.$$

Hint: Remember your givens.

Last edited:
Shooting Star
Homework Helper
how can i use the given 75% its max height and the 4/5 speed of the projectile

You only have to find the ratio of vx/vy_0, not the absolute values.

Thanks!! Got it!