At the top of a cliff 100 m high, Raoul throws a rock upward with a velocity of 15.0 m/s. How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff

Answer 1

We can use the kinematic equations of motion to solve this problem. We know that the acceleration due to gravity is 9.81 m/s^2 and that the distance traveled by both rocks is 100 m.

For the first rock, we can use the equation:

y = y0 + v0t + 1/2 at^2

where y, y0, v0, a, and t represent the final height, initial height, initial velocity, acceleration, and time, respectively.

Substituting in the values given:

100 = 0 + 15t - 1/2(9.81)t^2

Simplifying and rearranging:

4.905t^2 - 15t + 100 = 0

Using the quadratic formula:

t = (15 ± √(15^2 - 4(4.905)(100)))/(2(4.905))

t ≈ 6.42 s (time for the first rock to hit the ground)

For the second rock, we want it to hit the ground at the same time as the first rock, so we can use the equation:

y = y0 + 1/2 at^2

where y0 is the initial height (100 m).

Substituting in the values given:

100 = 0 + 1/2(9.81)t^2

Simplifying and solving for t:

t ≈ 4.48 s

Therefore, Raoul should drop the second rock 6.42 - 4.48 = 1.94 seconds after throwing the first rock.

Answer 2

AAAaannndd the bot gets it wrong yet again!

To find how long it takes the 1st rock to hit the ground, solve
100 +15t - 4.9t^2 = 0
t = 6.30

for the 2nd rock, solve
100 - 4.9t^2 = 0
t = 4.52

Therefore, Raoul should drop the second rock 6.30 - 4.52 = 1.78 seconds after throwing the first rock.

Answer 3

Apologies for the mistake in my previous response. You are correct, the time for the first rock to hit the ground is 6.30 seconds, and the time for the second rock to hit the ground is 4.52 seconds.

Therefore, Raoul should drop the second rock 6.30 - 4.52 = 1.78 seconds after throwing the first rock to ensure they hit the ground simultaneously. Thank you for bringing this to my attention and allowing me to correct my mistake.