# Projectile: launching and landing from the same height help me ?

1. Nov 22, 2007

### uaeXuae

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

If an athlete jumped 2 feet high and left the ground at an angle of 20 degrees with respect to the horizontal, how fast was the athlete going in the forward (positive horizontal) and upward (positive vertical) directions immediately after takeoff? If the height of takeoff was the same as the height of landing, how fast was the athlete going in the horizontal and vertical directions right before landing?

2. Relevant equations

vx = vcos(thata)
vy = vsin(theta)

Immediately after take off :
vx = 2cos25 = 1.8126
vy = 2sin25 = 0.845

right before landing:

vx = 2cos25 = 1.8126
vy = 2sin25 = -0.845

3. The attempt at a solution

Im not sure if my solution is correct can anyone confirm please ?

2. Nov 22, 2007

### rl.bhat

Angle is 20 not 25. What is the equation for h?

3. Nov 22, 2007

### uaeXuae

Sorry i put the wrong theta i took the theta for another problem.

The equation for H is as follows:

So basically what i think its going to be the 2nd one from the Y- Direction Equations?

4. Nov 22, 2007

### rl.bhat

You can use third one. In that Vfy = 0.

5. Nov 22, 2007

### uaeXuae

If i use the third one i will have one unknown which is d

dy = -viy^2/2a

dy = -(2sin20)^2/2(32)

dy = 7.3 * 10 ^ 03 ??

6. Nov 22, 2007

### rl.bhat

Here d=2 feet not vi

7. Nov 22, 2007

### uaeXuae

THANKSSS

Rearranging :

viy^2 = -2(-32)(2)
viy = 11.3 m/s

At take off:
Vv= 11.3 m/s

At landing:
Vv = -11.3 m/s

Heres the solution after writing it down , all coments are welcomed:

Last edited: Nov 22, 2007
8. Nov 22, 2007

### rl.bhat

Good!!

9. Nov 22, 2007

### uaeXuae

All Credits go to you thanx a lot