Question

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

Answer 1

The two stones hit the water 11.98 seconds after the release of the first stone. The initial velocity of the second stone must be 1.82 m/s for both stones to hit the water simultaneously. The speed of the first stone at the instant it hits the water is 108.334 m/s, and the speed of the second stone is 19.579 m/s.

Step-by-step explanation:

Given that the cliff height is 54.0 m and the first stone is thrown vertically downward with an initial speed of 1.82 m/s, we can calculate the time it takes for the first stone to hit the water using the equation for time:

t = (2 * h) / g

Where t is the time, h is the height, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values, we get:

t = (2 * 54.0) / 9.8

t = 10.98 s

Since the second stone is thrown 1.00 s after the first stone, we can calculate when it will hit the water:

t2 = t1 + 1.00

t2 = 10.98 + 1.00

t2 = 11.98 s

Therefore, the two stones hit the water 11.98 s after the release of the first stone.

To determine the initial velocity of the second stone for both stones to hit the water simultaneously, we can use the equation of motion:

v = u + gt

Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

Since both stones will hit the water at the same time, we can set their final velocities equal to each other and solve for the initial velocity of the second stone:

1.82 - 9.8 * t = 0

1.82 - 9.8 * (t1 + 1.00) = 0

t1 = 10.98

1.82 - 9.8 * (10.98 + 1.00) = 0

So the initial velocity of the second stone must be 1.82 m/s.

Finally, if we want to find the speed at which each stone hits the water, we can use the equation of motion again:

v = u + gt

For the first stone:

v1 = 1.82 + 9.8 * 10.98

v1 = 108.334 m/s

For the second stone:

v2 = 1.82 + 9.8 * (10.98 + 1.00)

v2 = 19.579 m/s

Answer 2

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

Step-by-step 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