Question

The ceiling of a long hall is 20m high what is the maximum horizontal distance that a ball thrown with a speed of 40m/s can go without hitting the ceiling of the hall (g = 10m/S2) ?

Answer 1

Given :

The ceiling of a long hall = 20m

Initial velocity = 40m/s

To find :

The maximum horizontal distance.

Solution :

First we have to find the angle of the ball thrown.

$: \implies \sf H = \dfrac{ {u}^{2} {sin}^{2} \theta}{2g}$

$: \implies \sf 20 = \dfrac{ {(40)}^{2} {sin}^{2} \theta}{2 \times 10}$

$: \implies \sf 20 = \dfrac{ 1600 \: {sin}^{2} \theta}{20}$

$: \implies \sf \dfrac{400}{1600} = {sin}^{2} \theta$

$: \implies \sf {sin}^{2} \theta = 0.25$

$: \implies \sf sin \theta = 0.5$

$: \implies \sf \theta = 30 \degree$

we know,

$: \implies \sf R = \dfrac{ {u}^{2}sin \: 2 \theta }{g}$

$: \implies \sf R = \dfrac{ {(10)}^{2}sin \: 60 \degree}{10}$

$: \implies \sf R = \dfrac{ 1600 \times 0.866 }{10}$

$: \implies \sf R = 138.56 \: m$

thus, the maximum horizontal distance is 138.56 m.