Question

a stone is thrown from the top of a tower of height 50m with a velocity of 30ms-1 at an angle of 30° above the horizontal. find the time during which the stone will be in air​

Answer 1

Given:

A stone is thrown from the top of a tower of height 50m with a velocity of 30ms-1 at an angle of 30° above the horizontal.

To find:

Time for which it's in air

Calculation:

Considering the Y axis velocity component of the projectile :

$\boxed{v_{y} = 30 \times \sin(30 \degree) = 15 \: m {s}^{ - 1} }$

Considering downward direction as positive :

$\therefore \: s = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: 50 = ( - 15)t + \dfrac{1}{2} g {t}^{2}$

$= > \: 50 = ( - 15)t + ( \dfrac{1}{2} \times 10 \times {t}^{2} )$

$= > \: 50 = ( - 15)t + 5 {t}^{2}$

Cancelling common terms by 5 :

$= > \: 10 = ( - 3)t + {t}^{2}$

$= > \: {t}^{2} - 3t - 10 = 0$

$= > \: {t}^{2} - (5 - 2)t- 10 = 0$

$= > \: {t}^{2} - 5t + 2t- 10 = 0$

$= > \: t(t - 5) + 2(t- 5) = 0$

$= > \: ( t + 2)(t - 5) = 0$

So , t - 5 = 0

=> t = 5 sec.

So , the ball will stay in air for 5 seconds.

Answer 2

Answer:

pls mark as brain list if it is correct

Explanation: