Question

A ball is thrown vertically upwards it goes to a height of 19.6m and then comes back. Find:- (i) Initial velocity of ball (ii) Total time of journey (iii) the final velocity of the ball when it serves the ground. g=9.8m/s^2

Answer 1

Answer:

(i) Initial Velocity = 19.6 m/s^2

(ii)Total time of journey = 4 sec

Explanation:

(i) Given that,

Height = $h$ = 19.6m

Acceleration due to gravity = $g$ = -9.8 $m/s^{2}$ (as the ball is thrown upward, we will take the acceleration due to gravity -ve)

Initial velocity = $v_{i}$ = ?

sol:

According to the equation of motion,

$2gh = v_{f} ^{2} - v_{i} ^{2}$

Substituting the values we get,

$2\times-9.8\times19.6 = 0 - v_{i}^2$

$v_{i}^2 = 384.9$

$v_{i} = 19.6 m/s^{2}$

(ii) Total Time of journey,

First we will find the time taken to reach at 19.6m by the ball,

accourding to the formula,

$v_{f} = v_{i} + gt$

Substituting the values,

$0 = 19.6 - 9.8t\\9.8t = 19.6\\t = 2 sec$

For the total time of the journey we need to muliply $t$ by 2,

$T = 2t\\T = 2 \times 2\\T = 4 sec$