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Question:
A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.
Answer:
(a) To find: The velocity with which it was thrown up.
Time of ascent is equal to the time of descent. The ball takes a total of 6s for its upward and downward journey.
Hence, time taken for the upward journey,
At maximum height final velocity
Using the first equation of motion:
The velocity with which the stone was thrown up is 29..4 m/s.
(b) To find: The maximum height ball reaches.
Using the second equation of motion
The maximum height ball reaches is 44.1 m.
(c) To find: Ball's position after 4 sec.
Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards.
In this case, Initial velocity,
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s - 3 s = 1 s
Distance travelled in another 1sec = s'
The distance travelled in another 1 sec = 4.9 m.
Therefore in 4 sec, the position of ball = (44.1 - 4.9) = 39.2 m from the ground.
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