Answer:
0.82 m
Explanation:
The initial velocity of the ball (horizontal velocity) is
[tex]u_x=101.0 mi/h \cdot \frac{1609 m/mi}{3600 s/h}=45.1 m/s[/tex]
And this horizontal velocity is constant during the motion of the ball.
The horizontal distance covered by the ball is
[tex]d=60.5 ft = 18.4 m[/tex]
So the time it took for the ball to reach the ground was
[tex]t=\frac{d}{v_x}=\frac{18.4 m}{45.1 m/s}=0.41 s[/tex]
The vertical distance covered by the ball during this time interval is
[tex]y=\frac{1}{2}at^2[/tex]
where
a = g = 9.8 m/s^2
is the acceleration due to gravity
Substituting,
[tex]y=\frac{1}{2}(9.8 m/s^2)(0.41 s)^2=0.82 m[/tex]
