Question

A ball is thrown upwards with a velocity of 29ms

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in the absence of

air resistance. A) Mention the direction of the acceleration during

the upward motion of the ball. B) What is the velocity and

acceleration of the ball at the highest point? C) How much high the

ball goes up? D) Find the time taken by the ball to reach the ground. g=9.8ms

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Answer 1

Answer:

Solution :

The direction of acceleration is in the downward direction i.e.,opposite to the direction of the motion of the ball. (ii) Velocity of the ball at the highest point = 0 Acceleration of the ball at the highest point = g = 9.8 ms−2 in the downwards direction

(iii) Here u=29.4ms−1,v=0 a=−g=−9.8ms−2

Let h= maximum height attained by the ball

using v2−u2=2gh,we have , 0−(29.4)2=−2×9.8h ∴ h=44.1m

(iv) Let time taken by the ball to reach the highest point = t Now, using we get, 0=29.4−9.8tort=3s

Since same time is taken by the ball to reach the ground from the highest point

∴ Time taken by the ball to reach the ground , 3s+3s=6s