# Baseball projectile motion question

[SOLVED] Baseball projectile motion question

## Homework Statement

A ball player hits a home run, and the baseball just clears a wall 8.00 m high located 104.8 m from home plate. The ball is hit at an angle of 38.4 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.2 m above the ground. The acceleration of gravity is 9.81 m/s^2.
a) What is the initial speed of the ball?
b) How much time does it take for the ball to reach the wall?
c.) Find the speed of the ball when it reaches the wall.

## Homework Equations

x(t)-Vo cos(theta)t
y(t)= -.5gt^2 +Vosin(theta)t+yo
V^2 = Vo^2 - 2 g (y - y°)

## The Attempt at a Solution

Ok this is what I did...
Let X axis horizontal, and Y axis vertical upward. The equations of the ball are:
x(t) = Vo cos38.4° t
y(t) = - (1/2) g t^2 + Vo sin38.4° t + yo.
When x(t) = 104.8 m , y(t) = 8 m.
Solving : Vo = 33.9 m/s (a) ; t = 3.94 s (b)
(c) Energy's conservation
V^2 = Vo^2 - 2 g (y - y°)
V = 31.88 m/s (c)

I'm not sure if this is right though. I would appreciate any help. I think you may have to compensate for the 1.2 m high starting position.

For part A. How did you get around your unknown: time?

Well, I have changed what I think it is since i posted.I tried using the SUVAT equations and I actually compensated for the 1.2 m initial height. See work below:
Horizontal: s=104.8, v=u=Ucos(38.4), t=?
Vertical: s=6.8, a=-9.81, u=Usin(38.4), t=?, v=0
s=vt-0.5a(t^2) (one of the "suvat" equations)
6.8/4.905=t^2

back to horizontal...
Ucos(38.4)=s/t=104/1.18 (speed=distance/time)
U=104/1.18cos(38.4)

v=112cos(38.4)
v=87.8 m(s^-1)

back to vertical: v=u+at=112sin(38.4)-9.81x1.18 (another suvat equation)
v=58.0 m(s^-1)

overall velocity= [58^2+87.8^2]^0.5 (pythagoras)