John drops a stone, from rest, at the top of a 20meter high cliff. His friend Bo

Question

John drops a stone, from rest, at the top of a 20meter high cliff. His friend Bob throws stone upward from the base of the cliff giving it an initial velocity of 30m/s. The stones are thrown/dropped at exactly the same instant. Assume Bob's stone is initially on the ground at a position of 0 meters, while John's stone starts 20 meters from the ground and that both stones experience only free fall motion.
a) What is velocity of John's stone 1s after he has dropped it?
b) At what time, does Bob's stone to hit the ground?
c) At what time, do the two stones pass each other?

Explanation / Answer

In this question we take the convention that "-ve" vectors are directed downwards. You've got to be careful of the signs in this question

(a) Kinematic Equation v= u + at = 0 - gt = -(9.81)(1) = -9.81 m/s

(b) Kinematic Equation s=ut+(1/2)at2 ---> 0= (30)t + (1/2)(-9.81)t2 (Note that the sign of the velocity and the acceleration are opposite because they are in different directions)

---> Solving for t = 6.11s

(c) Sj = (1/2)*(-9.81)t2 + 20 (Note the term +20m because john started 20m higher)

Sb = (30)t + (1/2)(-9.81)t2

Equating the 2 displacements: solving t= 2/3 s

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