Physics Homework

Question

A rock climber stands on top of a 50-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 2.0 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 each stone as it hits the water?

Answer 1

For the first stone:

The distance travelled is the initial velocity times time + distance for an acceleration from 0.

In this case, the acceleration is g (the acceleration of gravity--- 9.8 m/s² )

So

d = v_i * t + 1/2 * g * t²

Plug in the numbers:

50 = 2*t + 9.8/2 * t²

Put in standard format:

4.9t² + 2t - 50 = 0

not obviously factorable, so use the quadratic formula:

t = (-2 ± √( 2² - 4*4.9*(-50)) ) / 2(4.9)

t = (-2 ± √984 ) / 9.8

t = (-2 ± 31.4) / 9.8

A negative value has no meaning so t = 29.4/9.8 = 3 (This is the (a) answer)

This is the time after the 1st throw that BOTH stones hit the water, so the time for the second stone is 2 seconds.

Use the same equations to get the initial velocity of the 2nd stone. (This time, you'll be entering d and t and solving for v_i )

For the velocity hitting the water, use v = v_i + g*t = v_i + 9.8t

plug in v_i and t, and solve