Question

A ball is thrown straight upwards with an initial velocity 19.6 m/s. it was caught at the same distance above the ground from which it was thrown. 1) how high does ball rise ? (2)how long does the ball remains in air? ( g = 9.8ms-2)​

Answer 1

Explanation:

a) Maximum Height reached by the ball = u²/2g

= 19.6 × 19.6/2 × 9.8

= 19.6 m

b) Time ball remains in air

= √2h/g

= √2 × 19.6/9.8

= √39.2/9.8

= √4 = 2 seconds