A mountain climber stands at the top of a 33.5-m cliff that overhangs a calm poo

Question

A mountain climber stands at the top of a 33.5-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 -2.40 m/s. (Indicate the direction with the sign of your answers.) (a) How long after release of the first stone d d 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

(A) y = - 33.5 m

v0y = -2.40 m/s

ay = - 9.81 m/s^2

y = v0y t + ay t^2 /2

- 33.5 = - 2.40t - 4.9t^2

4.9 t^2 + 2.40t - 33.5 = 0

t = 2.38 sec ......Ans

(B) time for second stone, t = 2.38 - 1 = 1.38 sec

putting in equation,

- 33.5 = v(1.38) - (9.81 x 1.38^2 / 2)

v = -17.5 m/s

(C) first stone:

vf = vi + a t

v = -2.40 + (-9.8)(2.38)

v = - 25.7 m/s

Second Stone:

v = - 17.5 - (9.8 x 1.38)

v = - 31 m/s

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