Answer:
Part b)
h = 78.5 m
Part c)
v = 39.24 m/s
Explanation:
Part b)
If ball need t = 0 to t = 4 s then height of the tower is the total displacement of the ball in t = 4 s interval
here if ball start from rest
then its displacement is given as
[tex]\Delta y = \frac{1}{2}gt^2[/tex]
[tex]\Delta y = \frac{1}{2}(9.81)(4^2)[/tex]
[tex]\Delta y = 78.5 m[/tex]
Part c)
Speed of the bearing at the end of the motion of the ball
[tex]v_f = v_i + at[/tex]
[tex]v_f = 0 + (9.81)(4)[/tex]
[tex]v_f = 39.24 m/s[/tex]