Physics problem need help!!?

Question

A ball player hits a home run, and the baseball just clears a wall 8.00 m high located 132.8m from the home plate. The ball is hit at an angle of 38.8 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.1 m above the ground. The acceleration of gravity is 9.8 m/s^2.

What is the initial speed of the ball in units of m/s?

How much time does it take for the ball to reach the wall in units of seconds?

Find the speed of the ball when it reaches the wall. Answer in units of m/s

Answer 1

The equations of motion are
y(t)=1.1+v0*sin(38.8)*t-.5*9.8*t^2

x(t)=v0*cos(38.8)*t

when x(t)=132.8 and y(t)=8, the ball clears the wall

using x(t)
v0*t=132.8/cos(38.8)
=103.5
now use that is y(t)
8=1.1+103.5*sin(38.8)-.5*g*t^2

solve for t, which is the flight time
t=4.51 seconds

now solve for v0
37.7 m/s

The speed when clearing the wall is
vy=v0*sin(38.8)-9.8*t
vy=-20.6 m/s

vx=v0*cos(38.8)
vx=29.4 m/s

v=sqrt(vx^2+vy^2)
v=47.85 m/s

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