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 5 s. Calculate : (i) the height
of the tower, (ii) the velocity of ball on reaching the
ground. Take g = 9.8 m/s²​

Answer 1

Answer:

v = u + a × t

s=ut+ 21 at 2=19.6×5− 21 2 ×9.8×5 =−24.5 m

Hence, height of tower =24.5m

v=u+at

=19.6−9.8×5=−29.4 m/s

Answer 2

Answer:

s=ut+

2

1

at

2

=19.6×5−

2

1

×9.8×5

2

=−24.5 m

Hence, height of tower =24.5m

v=u+at

=19.6−9.8×5=−29.4 m/s