Question

A particle is projected from ground with velocity 20(sqrt2) m//s at 45^@. At what time particle is at height 15 m from ground? (g = 10 m//s^2)

Answer 1

Answer:

Given:

Initial Velocity of Projectile = 20√2 m/s

Angle of Projection = 45°

To find:

Time at which the Projectile is 15 m above the ground.

Concept:

We shall consider the y - axis component of Velocity . Treating the y axis Displacement like a linear motion, we can apply the equations of motions to find out the time taken.

Calculation:

Velocity in y axis :

$v_{y} = v \sin( \theta)$

$= > v_{y} = v \sin( 45 \degree)$

$= > v_{y} = \dfrac{v}{ \sqrt{2} }$

Now applying equation of motion :

$\therefore \: h = v_{y}t - \dfrac{1}{2} g {t}^{2}$

$= > 15 = \dfrac{20 \sqrt{2} }{ \sqrt{2} } t - (\dfrac{1}{2} \times 10 \times {t}^{2} )$

$= > 15 = 20t - 5 {t}^{2}$

Cancelling all terms by 5:

$= > {t}^{2} - 4t + 3 = 0$

$= > {t}^{2} - 3t - t + 3 = 0$

$= > t(t - 3) - 1(t - 3) = 0$

$= > (t - 1)(t - 3) = 0$

$= > t = 1 \: sec\: or \: 3 \: sec$

So final answer :

$\boxed{ \huge{ \red{ \bold{ \sf{ t = 1 \: sec\: or \: 3 \: sec}}}}}$

Answer 2

$\huge\bold\green{Question:-}$

A particle is projected from ground with velocity 20(sqrt2) m//s at 45^@. At what time particle is at height 15 m from ground? (g = 10 m//s^2).

$\huge\bold\green{Answer:-}$

Velocity in y axis = v sinθ

vy = v Sin45°

$vy = v \div \sqrt{2}$

We need to apply the equation of motion as:

h = vyt - 1/2gt^2

$15 = 20 \sqrt{2 \div \: \sqrt{20t} }$

$- 1 \div 2 \times 10 \times t {}^{2}$

$15 = 20t - 5t {}^{2}$

We get,

$t {}^{2} - 4t + 3 = 0$

$t {}^{2} - 3t - t + 3 = 0$

$t(t - 3) - 1(t - 3)$

$(t - 1)(t - 3) = 0$

$= > t = 1s \: or \: it \: can \: be \: 3s$