Question

A ball is thrown vertically upwards from the top of a tower at 4.9 ms–1. It strikes the pond near the base of the tower after 3 seconds. The height of the tower is [Take g = 9.8 m/s2]

Answer 1

Hello ★★


Just apply equations of motion and you will get your answer.


=> S= ut + ½(gt²)
S= height of tower
U= initial speed= -4.9m/s.

(Upward hence -ve.)T= time = 3sec .g= 9.8m/s².



→ On solving s= 29.4 m

Answer 2

The height of the tower is 29.4 m.

Explanation:

It is given that,

The initial speed of a ball from the top of a tower is 4.9 m/s

It strikes the pond near the base of the tower after 3 seconds.

We need to find the height of the tower. It is based on the concept of kinematics. The height of the tower is given by the formula as follows :

$h=ut+\dfrac{1}{2}gt^2$

Here, u = -4.9 m/s as the ball is thrown upwards

$h=-4.9\times 3+\dfrac{1}{2}\times 9.8\times (3)^2\\\\h=29.4\ m$

So, the height of the tower is 29.4 m.

Learn more,

Kinematics

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