A player hits a home run, and the base ball clears a wall 8 meters high located 113.6 m away from home plate.?

Question

The ball is hit at an angle of 38.7 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.2m above ground.
what is the initial speed of the ball? (m/s)
How much time does it take for the ball to reach the wall? (s)
Find speed of ball when it reaches the wall. (m/s)

Answer 1

The equations of motion are
y(t)=1.2+v0*sin(38.7)*t-.5*9.8*t^2

x(t)=v0*cos(38.7)*t

when x(t)=113.6 and y(t)=8, the ball clears the wall

using x(t)
v0*t=113.6/cos(38.7)
=145.6
now use that is y(t)
8=1.2+145.6*sin(38.7)-.5*g*t^2

solve for t, which is the flight time
t=4.146 seconds

now solve for v0
39.79 m/s


vy=v0*sin(38.7)-9.8*t
vy=-18.6731 m/s

vx=v0*cos(38.7)
vx=27.4 m/s

v=sqrt(vx^2+vy^2)
v=44.54 m/s

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