A ball is thrown with enough speed straight up so that it is in the air several

Question

A ball is thrown with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it gets to its highest point? (b) What is its velocity 2s before it reaches its highest point? m/s (c) What is the change in its velocity during this 2s interval? m/s (d) What is its velocity 2 s after it reaches its highest point? m/s (e) What is the change in velocity during this 2 s interval? m/s (f) What is the change in velocity during the 4 s interval? (Careful!) m/s (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?

Explanation / Answer

a] Highest point means that the ball cannot go any higher. This means that it momentarily comes to rest and so the velocity at the highest point is zero.

b] let t be the when the ball is at its highest point.

so, v = u - gt

=> 0 = u - 9.8t

=> t = (u/9.8) seconds

so, at (t - 2) seconds, the velocity will be:

v = u - 9.8(t - 2)

=> v = 9.8t - 9.8(t - 2) = 19.6 m/s. This is the velocity 2 seconds before reaching the highest point.

c]

At the highest point, v = 0 m/s

so change in velocity during this 2 s interval is 0 -19.6= - 19.6 m/s.

d] Velocity of the ball 2 seconds after it reaches the highest point will be: v = - 19.6 m/s [downwards].

e] Change in velocity in this 2 seconds will be: - 19.6 - 0 = - 19.6 m/s.

f] Change in velocity in the 4 seconds interval will be: 19.6 - ( - 19.6) = 39.2 m/s.

g] Acceleration of the ball at any time is a = g = - 9.8 m/s2.

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