Question

A rock climber stands on top of a 60 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s .

A. How long after the release of the first stone does the second stone hit the water?

B. What was the initial speed of the second stone?

C. What is the speed of the first stone as it hits the water

D. What is the speed of the second stone as it hits the water?

Answer 1

A) The distance traveled by the first stone is

This is a quadratic equation whose positive solution is

The second and first stone hit the water at the same time

Thus, the second stone hits the water 3.33 seconds after the release of the first stone

B) Time taken by the second stone is

Let the initial speed of second stone be u2

Thus the distance covered by the second stone is

Then the solution of this equation gives the initial speed of the second stone

C) Speed of the first stone as it hits the water is

D) speed of the second stone as it hits the water is