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

[uaeXuae]

Homework Statement

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?

Homework 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

The Attempt at a Solution

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

 

[rl.bhat]

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

 

[uaeXuae]

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

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

The equation for H is as follows:

http://aycu12.webshots.com/image/34651/2001284047037523326_rs.jpg

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

 

[rl.bhat]

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

 

[uaeXuae]

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

0 = viy^2 + 2ady

dy = -viy^2/2a


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

dy = 7.3 * 10 ^ 03 ??

 

[rl.bhat]

Here d=2 feet not vi

 

[uaeXuae]

THANKSSS

0 = viy^2 + 2ady

Rearranging :
Viy^2 = -2ady


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:

http://aycu37.webshots.com/image/32596/2000992234846000573_rs.jpg

 

[rl.bhat]

Good!

 

[uaeXuae]

All Credits go to you thanks a lot