Question

An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.02 m/s.

(a) How long after release of the first stone do the two stones hit the water?
s

(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?

magnitude	m/s
direction	---Select--- upward downward

(c) What is the speed of each stone at the instant the two stones hit the water?

first stone	m/s
second stone	m/s

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

a) Vo*t + 0.5g*t^2 = d.

1.92*t + 4.9*t^2 = 49.

4.9t^2 + 1.92t - 49 = 0.

Use Quadratic Formula and get:

t = 2.9714 seconds.

b. Vo*2.9714 + 4.9*2.9714^2 = 49.

Vo*2.9714 + 43.263 = 49.

2.9714*Vo = 49 - 43.263 = 5.7368.

Vo = 5.7368 / 2.9714 = 1.93068 m/s. ----> Intial Velocity

Direction Vertically Downward

c. Vf^2 = Vo^2 + 2g*d.

Vf^2 = 1.92^2 + 19.6*49 = 964.086.

Vf = 31.049 m/s. = Final velocity.