Question

An inquisitive physics student and mountain climber climbs a 48.0-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 2.18 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

Answer 1

Answer:

a) The two stones hit the water 2.91 s after the release of the first stone.

b) The initial velocity of the second stone must be 15.8 m/s vertically downward.

c) The velocity of each stone when they reach the water is:

First stone : -30.7 m/s

Second stone: -34.5 m/s

Explanation:

The height and velocity of the stones can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity at time "t".

a) If we place the origin of the frame of reference on the water, then, the height of both stones when they hit the water will be 0. Using the equation of height for the first stone, we can obtain the time when the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = 48.0 m -2.18 m/s · t - 1/2 · 9.81 m/s² · t²

0 = 48.0 m - 2.18 m/s · t - 4.91 m/s² · t²

Solving the quadratic equation:

t = 2.91 s

The two stones hit the water 2.91 s after the release of the first stone.

b) The second stone reaches the water in (2.91 s - 1.00 s) 1.91 s after released (remember that it was released 1.00 s after the first stone but both reached the water simultaneously). Then, using the equation of height, we can obtain the initial velocity knowing that at t = 1.91 s, y = 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = 48.0 m + v0 · 1.91 s - 1/2 · 9.81 m/s² · (1.91 s)²

(-48.0 m + 1/2 · 9.81 m/s² · (1.91 s)²) / 1.91 s = v0

v0 = -15.8 m/s

The initial velocity of the second stone must be 15.8 m/s vertically downward.

c) We have to use the equation of velocity for each stone. We already know the time when the stones reach the water and the initial velocities:

First stone:

v = v0 + g · t

v = -2.18 m/s - 9.81 m/s² · 2.91 s

v = -30.7 m/s

Second stone:

v = v0 + g · t

v = -15.8 m/s - 9.81 m/s² · 1.91 s

v = -34.5 m/s

The velocity of each stone when they reach the water is:

First stone : -30.7 m/s

Second stone: -34.5 m/s