A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm poo

Question

A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm pool of water.She throws two stones vertically downward 1.00 s apart and observesthat they cause a single splash. The first stone had an initialvelocity of -2.60 m/s. (a) How long after release of the first stonedid the two stones hit the water?
1 s

(b) What initial velocity must the second stone have had, giventhat they hit the water simultaneously?
2 m/s

(c) What was the velocity of each stone at the instant it hit thewater? first stone 3 m/s second stone 4 m/s (a) How long after release of the first stonedid the two stones hit the water?
1 s

(b) What initial velocity must the second stone have had, giventhat they hit the water simultaneously?
2 m/s

(c) What was the velocity of each stone at the instant it hit thewater? first stone 3 m/s second stone 4 m/s first stone 3 m/s second stone 4 m/s

Explanation / Answer

(a) How long after release of the first stone did the two stoneshit the water?        v ' = - [vo2 + 2 a h]    = - [-1.402 + 2 * 9.8 * 35 ]            = -26.14 m/s Therefore        t = v' - vo / a          = - 26.14 +1.40 / -9.8          = 2.525 s (b) What initial velocity must the second stone have had, giventhat they hit the water simultaneously?          vo' = h - 0.5 at'2 / t'              = -35 +0.5 * 9.8 * 1.0 / 1.0              = -30.1 m/s (c) What was the velocity of each stone at the instant it hit thewater?      final velocity of first stone v' = -26.14m/s      for second stone                v'' = vo ' + at'                    =-30.1 - 9.80 * 1                    = -39.9 m/s

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