A ball is thrown upwards with a velocity of `29.4 ms^(-1)` in the absence of air resistance.
(i) Mention the direction of the acceleration during the upward motion of the ball.
(ii) What are the velocity and acceleration of the ball at the highest point?
(iii) How much high the ball goes up?
(iv) Find the time taken by the ball to reach the ground. Take g = 9.8 `ms^(-2) `

Explanation

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= 29.4 ms^(-1) ,v=0 " "a =-g=-9.8 ms^(-2) `
Let h= maximum height attained by the ball
using `v^(2) -u^(2) =2gh , `we have , ` " "0-( 29.4 )^(2) =-2xx 9.8 h " "therefore " "h=44.1m`
(iv) Let time taken by the ball to reach the highest point = t Now, using we get, ` " " 0=29.4-9.8 t or t= 3s`
Since same time is taken by the ball to reach the ground from the highest point
` therefore " "` Time taken by the ball to reach the ground ,` " " 3s + 3s =6s `