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Question

Question 13
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.

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Solution

(i) According to the third equation of motion under gravity:
v2u2=2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s
During upward motion, g=9.8 ms2
Let h be the maximum height attained by the ball.
Hence,
0492=2×9.8×h
h=49×492×9.8=122.5 m

(ii) Let t be the time taken by the ball to reach the height 122.5 m, then according to the first equation of motion:
v = u + gt
We get,
0=49+t×(9.8)
9.8t=49
t=499.8=5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 s + 5 s = 10 s

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