Question

You drop a rock from rest out of a window on the top floor of a building, 25.0 m above the ground. When the rock has fallen 4.90 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the rock that your friend threw? (Assume upward is the positive direction and downward is negative. Indicate the direction with the sign of your answer.)

Answer 1

Answer:13.86 m/s

Explanation:

Given

building height=25 m

another rock is thrown after first rock has fallen 4.90 m

time taken by first rock to cover 4.90 m

$s=ut+\frac{gt^2}{2}$

$4.90=0+\frac{9.81\times t^2}{2}$

t=1 s

time taken by first rock to reach at bottom

$s=ut+\frac{gt^2}{2}$

$25=0+\frac{9.81\times t^2}{2}$

t=2.25 sec

Now second rock will take 1 s less than first rock

time for second rock=1.25 s

$s=ut+\frac{gt^2}{2}$

$25=u\times 1.25+\frac{9.81\times 1.25^2}{2}$

$u\times 1.25=25-\frac{9.81\times 1.25^2}{2}$

u=13.86 m/s