A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws two stones vertically downward 1.0s apart and observes that they cause a single splash. The initial speed of the first stone was 2.0m/s.

Well since they made a single splash, they hit the water at the same time and therefore the time after the release of the first stone until the second stone hits the water is also the time it takes the first stone to hit the water after the first stones release.

0 = -4.9t^2 2t 50

t=3 (time after first stones release until both stone hit the water)

so it only took the second stone 2s to hit the water after its own release since 3-1=2

0 = -4.9 x 4 2v 50

v = -15.2 (initial velocity of second stone)

For c part just multiply the respective times by acceleration and add to the initial velocities of the stones.

So for the first stone

v(final) = -2 9.83 = -31.4

and for the second stone

v(final) = -15.2 9.82 = -34.8

distance = d/v

use the formula speed = v/d, then rearrange to find out the time so it will be:

distance = d/v

This post is last updated on hrtanswers.com at Date : 1st of September – 2022