A mountain climber stands at the top of a 48.5-m cliff that overhangs a calm poo

Question

A mountain climber stands at the top of a 48.5-m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash. The first stone had an initial velocity of -1.40 m/s. (Indicate the direction with the sign of your answers.) (a) How long after release of the first stone did the two stones hit the water? (Round your answer to at least two decimal places.) s (b) What initial velocity must the second stone have had, given that they hit the water simultaneously? m/s (c) What was the velocity of each stone at the instant it hit the water? first stone m/s second stone m/s

Explanation / Answer

as stone reaches water surface,its vertical dispalcement will be -48.5 m.

and y = v0 t + a t^2 /2

both reaches at same time hence time of fall for first stone is t then time of fall for 2nd stone will be (t - 1).

y1 = -1.40 t - 9.8 t^2 / 2 = - 48.5

4.9 t^2 + 1.40t - 48.5 = 0

t = 3 sec

time of fall for second stone = 3 - 1= 2 sec

putting in,

- 48.5 = 2v0 - 4.9(2^2)

- 48.5 = 2 v0 - 19.6

v0 = -14.5 m/s

(A) t = 3 sec

(B) v0 = - 14.5 m/s

(C) v1 = -1.40 - 9.8(3) = - 30.8 m/s

v2 = -14.5 - 9.8(2) = - 34.1 m/s

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