An inquisitive physics student and mountain climber climbs a 46.0-m-high cliff t

Question

An inquisitive physics student and mountain climber climbs a 46.0-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 2.12 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 m/s direction (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

Explanation / Answer

(a)

we have the equation of motion

s =u*t+1/2*a*t2

for the first stone

u = 2.12 m/s

s = 46m

46 = 2.12*t + 1/2*9.8*t2

4.9*t2 + 2.12*t -46 = 0

t = 2.85 s

so time taken by the two stones to hit the water after thr releas of first stone is 2.85 s.

(b)

second stone is released after one second of first one. so time taken by the second stone to reach the water is 1.85 s.

we have equation of motion

s = u*t +1/2*a*t2

u = ( 2*s - a*t2 ) / 2*t

u = 2*46 - 9.8*1.85*1.85 / 2*1.85

u = 15.8 m/s

given that the stone thrown vertically downward .

so the magnitude and the direction of the initial velocity of second stone are 15.8 m/s downward .

(c)

final speedd of first stone

v1 = u1+a*t1

v1 = 2.12 + 9.8 * 2.85

v1 = 30.05 m/s

final velocity of second stone

v2 = u2 + a*t2

v2 = 15.8 + 9.8*1.85

v2 = 33.93 m/s

hense ,speed of first and second stone when it hit the water respectively are

30.05 m/s and 33.93 m/s .

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