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.1.78 sb.3.76 sc.2.48 sd.5.05 s
Question
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.1.78 sb.3.76 sc.2.48 sd.5.05 s
Solution
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)
Since the square root of a negative number is not a real number, we can conclude that the first rock will not hit the ground. This is because the initial upward velocity of the rock is greater than the velocity needed to overcome the gravitational pull.
Therefore, Raoul should drop the second rock immediately after throwing the first one, so they both hit the ground at the same time.
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