An inquisitive physics student and mountain climber climbs a 54.0-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 1.82 m/s.

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

(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

Question

Physics

Vivian Craig

7 June, 14:37

An inquisitive physics student and mountain climber climbs a 54.0-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 1.82 m/s.

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

(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

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Answers (1)

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Savannah · Answer 1

a) After the release of the first stone the two stones hit the water at 3.14 seconds.

b) Initial velocity of second stone = 14.74 m/s

c) Speed of first stone at the instant the two stones hit the water = 32.62 m/s

Speed of second stone at the instant the two stones hit the water = 35.73 m/s

Explanation:

a) For first stone we have

Initial velocity, u = 1.82 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 54 m

We have equation of motion s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

54 = 1.82 x t + 0.5 x 9.81 x t²

4.905 t² + 1.82 t - 54 = 0

t = 3.14 s or t = - 3.51 s (not possible)

After the release of the first stone the two stones hit the water at 3.14 seconds.

b) The second stone is thrown after 1 s

Time taken by second stone to reach top of pool of water = 3.14 - 1 = 2.14 s

Time, t = 2.14 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 54 m

We have equation of motion s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

54 = u x 2.14 + 0.5 x 9.81 x 2.14²

u = 14.74 m/s

Initial velocity of second stone = 14.74 m/s

c) We have equation of motion v = u + at

For first stone

Initial velocity, u = 1.82 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3.14 s

v = 1.82 + 9.81 x 3.14 = 32.62 m/s

For second stone

Initial velocity, u = 14.74 m/s

Acceleration, a = 9.81 m/s²

Time, t = 2.14 s

v = 14.74 + 9.81 x 2.14 = 35.73 m/s

Speed of first stone at the instant the two stones hit the water = 32.62 m/s

Speed of second stone at the instant the two stones hit the water = 35.73 m/s