Question

while standing at the edge of a cliff 45m high, you drop a ball. When the ball has fallen 5m, you throw a second ball straight down. What initial speed must you give the second ball if they are both to reach the ground at the same time?​

Answer 1

The initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.

Speed of the balls

The initial speed of the second ball is calculated as follows;

v = √2gh

v = √(2 x 9.8 x 45)

v = 29.7 m/s

Time for the ball to travel 5 meters

t = √(2h/g)

t = √(2 x 5/9.8)

t = 1.01 s

Initial speed of the first ball

vf = vi + gt

29.7 = vi + 9.8(1.01)

29.7 = vi + 9.9

vi = 19.8 m/s

Thus, the initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.

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