Question

20. A ball is thrown vertically upwards from the top
of a tower with an initial velocity of 19.6 m .
The ball reaches the ground after 5s. Calculate
(1) the height of the town the velocity of ball
on reaching the ground. Take g-9.8m​

Answer 1

Answer:

solved

Explanation:

height H covered and time T when it returns to the point of projection

H = u²/2g = 19.6²/19.6 = 19.6 m

T = 2u/g = 2x19.6/9.8 = 4 s

height for the remaining 1 second =ut + gt²/2 = 19.6 + 9.8/2 = 24.5 m

height of the tower = 24.5 m

the velocity of ball( when it reached the ground ) = 19.6 + 9.8 = 29.4 m/s