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 velocity of the second ball is 14.85 m/s

Motion Under Gravity

Motion under gravity is also known as vertical linear motion. The three formula involved are

• V = U +/- gt

• h = ut +/- 1/2g$t^{2}$

• $V^{2}$ = $U^{2}$ +/- 2gH

Where

• V = final velocity

• U = initial velocity

• g = acceleration due to gravity

• t = time

Given that 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.

Let first calculate the time for the first ball.

where U = 0

h = ut +/- 1/2g$t^{2}$

h = 1/2g$t^{2}$

substitute all the parameters

45 = 1/2 x 9.8 x $t^{2}$

45 = 4.9$t^{2}$

$t^{2}$ = 45/4.9

$t^{2}$ = 9.18

t = $\sqrt{9.18}$

t = 3.03 s

To calculate initial speed that you must give the second ball if they are both to reach the ground at the same time, we will make g negligible.

U  = h/t

U = 45/3.03

U = 14.85 m/s

Therefore, the initial velocity of the second ball is 14.85 m/s

Learn more about vertical motion here: https://brainly.com/question/24230984

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