Question

A ball thrown up is caught back by thrower after 6 sec. Calculate the velocity with which the ball was thrown up and maximum height attained by ball.

Answer 1

1.

Let ‘u’ be the initial velocity. The final velocity just before hitting the ground will be ‘-u’ (downward).

Using, v = u + at

=> -u = u – gt

=> u = gt/2

=> u = 9.8 × 6/2

=> u = 29.4 m/s

2.

Let, h be the maximum height reached. The time taken is 3 s.

h = ut – ½ gt²

=> h = 29.4 × 3 – ½ × 9.8 × 3²

=> h = 44.1 m