Answer:
Equations of motion of baseball are y(t) = h - \frac{g t^2}{2}y(t)=h−
2
gt
2
, x(t) = v_0 tx(t)=v
0
t, where hh is the initial height, and v_0v
0
is the initial horizontal velocity.
When the ball reaches the ground, y(t) = 0y(t)=0, from where it takes t_1 = \sqrt{\frac{2 h}{g}} \approx 0.55 st
1
=
g
2h
≈0.55s to reach the ground.
Substituting the time t_1t
1
from the last equation into equation for x(t)x(t), obtain the horizontal distance covered: s = x(t_1) = v_0 \sqrt{\frac{2 h}{g}} \approx 11.06 ms=x(t
1
)=v
0
g
2h
≈11.06m
