A home run is hit such a way that the baseball just clears a wall 22m high locat

Question

A home run is hit such a way that the baseball just clears a wall 22m high located 129m from home plate. The ball is hit at an angle of 40 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 2m above the ground. The acceleration of gravity is 9.8 m/s^2. What is the initial speed of the ball?
Knowns:
Delta X= 129m
Delta Y= 22-2 m = 20m
Vxi= Vicos40
Vxf= Vicos40
ax= O
ay= -9.8 m/s^2
Applicable Equation: Vxf^2= Vxi^2 + 2ax(Delta X)
I think this is how I am supposed to start but I am so lost at what to do...any tips?

Explanation / Answer

Initial speed is v

Vxo=vcos40

Vyo=vsin40

Horizontal displacement is x =129 m in t seconds

x = Vx0 * t

x = vcos 40 *t

vt = 129/cos 40

In same time t, Vertical displacement is

y=22-2=20

y=Vyot+1/2at^2

20=vsin40t-4.9t^2

4.9t^2 = sin 40 * (129/cos40)-20=88.24

t^2=88.24/4.9=18.008

t=4.24 m/s

vt=129/cos40

v=129/(cos40*4.24)=39.72 m/s

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