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 2.2 m/s .

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

What was the initial speed of the second stone?

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

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

Please explain how to solve the equations

Answer 1

mountain climber stands at the top of a 20.0 m cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The first stone has an initial velocity of 2.2 m/s.

The 2nd stone must travel the50.0 m downward in 1 second less than the 1st stone. Since both stones fall downward, I choose downward as positive direction.

Initial velocity = +2.2 m/s
Acceleration = +9.8 m/s^2
Displacement = +50 m

Let’s determine the final velocity of the 1st stone and the time of drop.
Vf^2 = Vi^2 + 2 * 9.8 * d
Vf^2 = 2.2^2 + 2 * 9.8 *50
Vf = 31.38 m/s = velocity of 1st stone when hit the water.

Time = (Vf – Vi) ÷ a
Time = (31.38 – 2.2) ÷ 9.8
Time = 2.97seconds for 1st stone to fall.

2nd stone is dropped 1 second later.
Time for 2nd stone to fall 50 m = 1.97 seconds

Displacement = (Initial velocity * time) + (1/2 * 9.8 * time^2)
50 = (Initial velocity * 1.97) + (1/2 * 9.8 * 1.97^2)
Initial velocity = [50 – (1/2 * 9.8 * 1.97 ^2)] ÷ 1.97 = 15.72 m/s

Final velocity = Initial velocity + (9.8 * time)
Final velocity = 15.72+ (9.8 *1.97) = 35 m/s


(a) How long after release of the first stone will the two stones hit the water? 2.987 seconds

(b) What initial velocity must the second stone have if they are to hit simultaneously?
15.72 m/s

(c) What will the velocity of each stone be at the instant they hit the water?

19.835 m/s (first stone)
35 m/s (second stone)