Question

A ball is projected horizontally with velocity 4 m/s from the top of a building 19.6 m high. The distance of the point from the bottom of the building where ball will hit the ground is?
(g =9.8 ms-2)

Answer 1

A ball is projected horizontally with a velocity of 4m/s from the top of a building 19.6m high. How long will the ball take to hit the ground.

The time taken to hit the ground is given by,

T= √2H / ​g

T = √2×19.6 / 9.8

T = 2s

please mark me brain mark list

​

Answer 2

Given:

A ball is projected horizontally with velocity 4 m/s from the top of a building 19.6 m high.

To find:

Distance at which the ball strikes ground?

Calculation:

This is an example of AIR-GROUND projectile:

$\therefore \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}$

$\implies \: 19.6 = 0 + \dfrac{1}{2} \times 9.8 \times {t}^{2}$

$\implies \: {t}^{2} = 4$

$\implies \: t = 2 \: sec$

Now, range (R) will be :

$R = u_{x}(t)$

$\implies \: R = 4 \times 2$

$\implies \: R = 8 \: metres$

So, range of projectile is 8 metres