Question
A rock is thrown horizontally with a speed of 20 m/s from a vertical cliff of height 25 m.
What is the velocity (magnitude and direction) of the rock just before it hits the ground?
What is the velocity (magnitude and direction) of the rock just before it hits the ground?
Answers
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Answered by
GPT 3.5
To find the velocity of the rock just before it hits the ground, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration (in this case, due to gravity, which is -9.81 m/s^2)
s = vertical distance (25 m)
Given that the initial velocity is horizontal, it does not contribute to the calculation. So, we can calculate the final velocity just before impact using the vertical distance the rock falls:
v = sqrt(2 * 9.81 * 25)
v = 31.3 m/s
Therefore, the magnitude of the velocity of the rock just before it hits the ground is 31.3 m/s. Since the rock is falling vertically at this point, the direction of the velocity will be downwards.
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration (in this case, due to gravity, which is -9.81 m/s^2)
s = vertical distance (25 m)
Given that the initial velocity is horizontal, it does not contribute to the calculation. So, we can calculate the final velocity just before impact using the vertical distance the rock falls:
v = sqrt(2 * 9.81 * 25)
v = 31.3 m/s
Therefore, the magnitude of the velocity of the rock just before it hits the ground is 31.3 m/s. Since the rock is falling vertically at this point, the direction of the velocity will be downwards.
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