a) Calculate the highest point in the ball path and answer if the person on the

Question

a) Calculate the highest point in the ball path and answer if the person on the cliff has any chances of catching the ball?

b) If so, at what time(s) the person may have a chance to catch the ball? (time measured from when the ball first thrown)

c) What are the velocity(s) of the ball at those moments?

d) If the person on the cliff fails to catch the ball, how long does it take for the ball to hit the ground?

e) In this case, calculate the average speed and velocity for the round trip of the ball, namely starting from ground level and then back to the ground level.

Explanation / Answer

part A: Hmax = Vo^2/2g

Hmax = 38*38/(2 * 9.81)

Hmax = 73.59 m , yes, there are chnaces of catching it

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time t = 2h/g

t^2 = 2* 60/9.81

t = 3.5 secs

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V = u +at

V = 38 -(9.8 * 2.5)

V = 13.5 m/s

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time of ascent = time of descent = 3.5 secs

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Vavg for the round trip = 0

Vavg speed = 38 + 13.5/2 = 25.75 m/s

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