Question

A ball is thrown upwards from the top of tower of height 80m with a velocity 20 m/s. Find the distance and displacement afte 5 seconds and it's velocity when the ball hits the ground?

Answer 1

Let's take g as 10m / s²

We will use the following formula :

S = ut - 0. 5gt²

Time taken to reach max height is :

u / g = 20 / 10 = 2 seconds.

In 2s the distance is :

Maximum height = 400 / 10 =40m

5 - 2 = 3 seconds.

The body will be falling.

U = 0

In 3 seconds S = 0.5 × 10 × 9 = 45 m

The distance in 5 seconds is :

45 + 40 = 85 m

Displacement in 5 seconds is :

45 - 40 = 5m

V is :

V² = u² + 2gs

S = 40 + 80 = 120

U = 0

V² = 2 × 10 × 120

V² = 2400

V = 48.989 m / s

48.99 m / s

Answer 2

Explanation:

$\huge \tt48.99m {s}^{ - 1} \: \: \sim \: \: 49m {s}^{ - 1}$