# Homework Help: Trajectory of diver with vertical and horizontal velocities

1. Sep 8, 2004

### Fused

Please help me with this problem! :yuck: :surprised I can't seem to get it set up quite right.
A high diver leaves the end of a 5 m high diving board and strikes the water 1.3s later 3 m beyond the end of the board. Considering the diver as a particle,d etermine a) her initial velocity, B. the maximum height reached and c. the velocity withwhich she enters the water.
I tried using these formulas: Vx = Vx0, x=x0+Vx0t
Vy = Vy0-gt, y = y0 + Vy0t- .5gt^2, v^2y = V^2y0 -2g(y-y0)

2. Sep 8, 2004

### Atheist

Hint: In this case movement in x-direction and in y-direction are independent.

Initial position (x0, y0) is given, vx0 is easy to compute, vy0 also computable. Knowing x0,y0,vx0,vy0 and the movement equation is sufficient to calculate the whole path. The formulas you present above are sufficient to solve the problem (still trying to find out what the last one is supposed to be - it can´t be correct just from looking at the units).

3. Sep 8, 2004

### BobG

Since the info provided only gave positions, I used the equation for position rather than velocity.

$$s_f=s_i+vt+\frac{1}{2}at^2$$
Solve for v (velocity)

Hint: You use the water as your frame of reference, making your final position 0. Acceleration due to gravity is downward, or negative.

That gives you your initial vertical velocity (or along the y-axis). Your horizontal velocity is just the change in horizontal position divided by time.

You can express your final answer in Cartesian coordinates (your horizontal velocity, vertical velocity) or polar coordinates (magnitude of the velocity vector, angle of trajectory). The magnitude of the vector is found by plugging the x-axis and y-axis velocities into the Pythagorean theorem.

4. Sep 9, 2004

### Fused

I think I get it now. Thanks