Why Do Two Stones Thrown Off a Cliff Make a Single Splash?

[fantolay]

Homework Statement

A person climbs 50m cliff that overhangs calm pool of water. He throws two stones vertically downward, 1s apart, and observes that they cause a single splash. The first stone has an initial speed of 2m/s downward.

(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 they are to hit simultaneously?
(c) What is the speed of each stone at the instant the two hit the water?

Homework Equations

Probably big 6 mechanics. We will have to calculate how long it takes the first stone to hit water then subtract one to get part a.
Once we figure out T we can probably figure out part b.
Not sure about part c.

 

[kuruman]

For part c, once you have the time of flight for each stone, just use the kinematic equation giving the velocity at any time t to find the final speed of each stone.

 

[kerimek]

I would approach this problem by first identifying the relevant equations from the big 6 mechanics: kinematic equations. These equations can help us calculate the time it takes for the first stone to hit the water and the initial velocity needed for the second stone to hit the water simultaneously.

For part (a), we can use the equation d = v0t + 1/2at^2, where d is the distance traveled (in this case, 50m), v0 is the initial velocity (2m/s), a is the acceleration due to gravity (9.8m/s^2), and t is the time. We can rearrange this equation to solve for t and plug in the given values to find that it takes approximately 3.3 seconds for the first stone to hit the water. Since the second stone is released 1 second after the first, it will hit the water 2.3 seconds after the first stone is released.

For part (b), we can use the same equation, but this time we know the initial velocity (2m/s) and the time (2.3 seconds). We can solve for the acceleration and find that the second stone needs an initial velocity of approximately 4.6m/s downward to hit the water at the same time as the first stone.

For part (c), we can use the equation vf = v0 + at, where vf is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. Since we know the initial velocity (2m/s), the acceleration (9.8m/s^2), and the time (2.3 seconds), we can solve for the final velocity and find that both stones will hit the water with a speed of approximately 22.5m/s.

In conclusion, using the kinematic equations from the big 6 mechanics, we can calculate the time it takes for the stones to hit the water, the initial velocity needed for the second stone to hit simultaneously, and the speed of each stone at the instant they hit the water.