A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. a. How long does it take to reach the ground? b. How far does the rock land from the base of the cliff?
Added by Stephen A.
Step 1
8 m/s^2 Using the equation: y = v0y * t + (1/2) * g * t^2 -20 = 0 * t + (1/2) * (-9.8) * t^2 -20 = -4.9t^2 t^2 = 20 / 4.9 t^2 = 4.08 t = √4.08 t ≈ 2.02 seconds Therefore, it takes approximately 2.02 seconds for the rock to reach the ground. Show more…
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