Question

A mountain climber stands at the top of a 41.5-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 answer.)

(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

Answer 1

Using 3rd kinematic equation for 1st stone:

v1^2 = u1^2 + 2*a*d

u1 = initial speed of 1st stone = -1.40 m/sec

We've assumed that downward direction is negative, So

a = -9.81 m/sec^2

d = vertical distance of pool = -41.5 m

v1 = sqrt (u1^2 + 2*a*d)

v1 = sqrt ((-1.40)^2 + 2*(-9.81)*(-41.5)) = -28.57 m/sec

Now Using 1st kinematic equation:

v1 = u1 + a*t1

t1 = time taken by 1st stone to hit the water = ?

t1 = (v1 - u1)/a

t1 = (-28.57 - (-2.5))/(-9.81) = 2.66 sec

Since given that they both hits at water together, So after 2.66 sec 2nd stone hits the water

Part B.

Now given that 2nd stone is released after 1.00 sec, So

t2 = traveling time of 2nd stone = 2.66 - 1.00 = 1.66 sec

Using 2nd kinematic equation for 2nd stone

h = U2*t2 + (1/2)*a*t2^2

U2 = (h - (1/2)*a*t2^2)/t2

Using known values:

U2 = ((-41.5) - 0.5*(-9.81)*1.66^2)/1.66

U2 = Initial speed of 2nd stone = -16.86 m/sec

U2 = -16.9 m/sec (where -ve sign indicates the direction of 2nd rock which is downward)

Part C.

From part A.

V1 = -28.57 m/sec = Velocity of 1st stone when it hits water

Part D.

Using 1st kinematic equation:

V2 = U2 + a*t2

V2 = -16.9 + (-9.81)*1.66

V2 = -33.18 m/sec

V2 = Velocity of 2nd stone after it hits water = -33.18 m/sec

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