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 2.2 m/s .

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

2.) What was the initial speed of the second stone?

3.) What is the speed of the first stone as it hits the water?

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

Answer 1

Answer 2

The time it takes for the first stone to hit the water can be found using the equation of motion for constant acceleration:

h = 1/2 gt^2

where h is the height of the cliff, g is the acceleration due to gravity (9.81 m/s^2), and t is the time it takes for the stone to hit the water.

Plugging in the given values, we get:

60 m = 1/2 (9.81 m/s^2) t^2

Solving for t, we get:

t = sqrt(120/9.81) = 3.91 s

Since the second stone was thrown 1.0 s after the first stone, it will hit the water 1.0 s later than the first stone:

t2 = t1 + 1.0 s = 4.91 s

The initial speed of the second stone can be found using the time it takes to hit the water and the distance it falls:

d = 1/2 gt^2

where d is the distance the stone falls and t is the time it takes to hit the water.

Plugging in the values, we get:

d = 1/2 (9.81 m/s^2) (4.91 s)^2 = 120 m

Since the distance fallen is the same for both stones, we can use the same equation to find the initial speed of the second stone:

v2 = sqrt(2gd) = sqrt(29.81120) = 38.2 m/s

The speed of the first stone as it hits the water can be found using the equation of motion for constant acceleration:

v = gt

where v is the final velocity of the stone (i.e., the speed as it hits the water).

Plugging in the values, we get:

v1 = (9.81 m/s^2) (3.91 s) = 38.3 m/s

The speed of the second stone as it hits the water can be found using the same equation as for the first stone:

v2 = (9.81 m/s^2) (4.91 s) = 48.2 m/s