Question

a ball thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m/s^-1 ,the ball reaches the ground after 5s calculate the height of the tower and velocity of ball on reaching the ground. Take g=9.8 m/s ^-2
please answer with steps ​

Answer 1

Answer:

Height of the tower = 24.5 m

Velocity of the ball = -29.4 m/s

Explanation:

s=ut+  1/2 at^2

 =19.6×5−  1/2 ×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