An inquisitive physics student and mountain climber climbs a 55.0-m-high cliff t

Question

An inquisitive physics student and mountain climber climbs a 55.0-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 2.16 m/s.

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

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

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

Explanation / Answer

a)

here by using the second equation of motion

y = u*t + 0.5 * a * t^2

55 = 2.16 * t + 0.5 * 9.8 * t^2

4.9t^2 + 2.16t - 55 = 0

then by soving the quadratic equation we get

t = 3.137 sec

b)

here also we use the second equation of motion

there is 1 sec less for the second stone

y = u*t + 0.5 * a *t^2

55 = u * ( 3.137 - 1 ) + 0.5 * 9.8 * ( 3.137 - 1)^2

55 = u * 2.137 + 4.9 * 2.137^2

u = 15.26 m/s

c)

use the first equation of motion

v = u + a * t

for first stone

v1 = 2.16 + 9.8 * 3.137

v1 = 32.9 m/s

then for second stone

v2 = 15.26 + 9.8 * 2.137

v2 = 36.2 m/s

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