At the top of a cliff 100 m high, Raoul throws a rock upward with velocity 20.0 m/s. How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff?Select one:a.2.48 sb.1.78 sc.3.76 sd.5.05 s

First, we need to find out how long it takes for the first rock to hit the ground. We can use the equation of motion:

h = ut + 0.5gt^2

where: h = height = 100 m u = initial velocity = 20 m/s g = acceleration due to gravity = 9.8 m/s^2 t = time

Rearranging the equation to solve for t gives us:

t = sqrt((2h - u^2) / g)

Substituting the given values into the equation gives us:

t = sqrt((2*100 - 20^2) / 9.8) = sqrt((200 - 400) / 9.8) = sqrt(-200 / 9.8) = sqrt(-20.41)

Since time cannot be negative, we discard this solution.

The other solution comes from the equation:

t = (u + sqrt(u^2 + 2gh)) / g

Substituting the given values into the equation gives us:

t = (20 + sqrt(20^2 + 29.8100)) / 9.8 = (20 + sqrt(400 + 1960)) / 9.8 = (20 + sqrt(2360)) / 9.8 = (20 + 48.59) / 9.8 = 7.01 s

This is the time it takes for the first rock to hit the ground.

The second rock is dropped from rest, so its initial velocity is 0. The time it takes to hit the ground can be found using the equation:

h = 0.5gt^2

Rearranging to solve for t gives us:

t = sqrt(2h / g)

Substituting the given values into the equation gives us:

t = sqrt(2*100 / 9.8) = sqrt(200 / 9.8) = sqrt(20.41) = 4.52 s

This is the time it takes for the second rock to hit the ground.

To find out when Raoul should drop the second rock, we subtract the time it takes for the first rock to hit the ground from the time it takes for the second rock to hit the ground:

4.52 s - 7.01 s = -2.48 s

Since time cannot be negative, we take the absolute value, which gives us 2.48 s.

Therefore, Raoul should drop the second rock 2.48 seconds after he throws the first rock for both rocks to hit the ground at the same time.

So, the answer is a. 2.48 s.