Question

A ball is thrown vertically upwards from *a tower* with a velocity of 19.6 m/s it takes 5 seconds to hit the ground calculate the height of the tower and the velocity with which the ball will hit the ground​

Answer 1

The ball is thrown upwards with velocity 19.6 m/s. During the upward motion it experiences -9.8 m/s² acceleration due to which it comes to rest momentarily at the highest point in air. We can calculate the time taken to reach the heighest point.

v = u + at

0 = 19.6 - 9.8 t

t = 2 sec

So the ball reaches the topmost point in air in 2 seconds.

Distance travelled by the ball until it reaches the highest point :

s = ut + at²/2 = 19.6 ×2 + 9.8 × 2²/2

s = 19.6×2 + 19.6 = 19.6×3 = 58.8 m

Hence the ball travels 58.8 m above the

height of tower after throwing.

Now the ball comes down and experiences an acceleration of +9.8 m/s². The time in which it reaches down from the highest point is 4 sec (6-2) because 2 sec is consumed in reaching the highest point.

Now let us calculate the distance travelled by the ball to reach the earth in 4 sec.

s = ut + at²/2

s = 0×t +9.8× 4²/2

s = 9.8× 8 m

This distance also includes the distance from the throwing point to the highest point, ie 58.8m. So we need to subtract that distance from this calculated distance of 9.8×8 m.

So height of tower = 9.8× 8 - 58.8

= 9.8×8 - 9.8×3

= 9.8×5

= 49 m.

Hence height of the tower is 49m