Question

A rock is dropped from a height of 110 meters above the ground. One second later, a ball is thrown vertically downward from the same height with an initial speed of 14 m/s.

A. At what time after the ball if thrown are the two objects at the same height above the ground?

B. What is the speed of the rock when the ball hits the ground?

Answer 1

A)
At a particular time, the distance travelled by the ball and rock will be the same.
Distance travelled by the rock in time t, Sr = ut + 1/2gt²
Where u is the initial velocity of the rock, u = 0
Sr = 1/2gt²
Distance travelled by the ball in time (t-1) s, Sb = 14 x (t-1) + 1/2 g(t-1)²
= 14t - 14 + 1/2 gt² - gt + 1/2g
= (14 - g) t - (14 - 1/2g) + 1/2 gt²

Sr = Sb
1/2gt² = (14 - g) t - (14 - 1/2g) + 1/2 gt²
(14 - g) t = (14 - 1/2g)
t = (14 - 1/2g)/(14 - g)
= (14 - 4.905) / (14 - 9.81)
= 2.17 s

B)
Initial velocity of the ball, u = 14 m/s
Distance travelled by the ball, s = 110 m
Acceleration, g = 9.81 m/s²

Using the formula, s = ut + 1/2 gt²,
0.5 x 9.81 t² + 14t - 110 = 0
Solving this quadratic equation, t = 3.52 s

Initial velocity of the rock, vi = 0
Acceleration, g = 9.81 m/s²
Time taken, t = 3.52 s
Using the formula, vf = vi + gt,
vf = 0 + 9.81 x 3.52
= 34.52 m/s