Question

If angle of projection is30 degree and velocity isroot 5 i cap +j cap then what will be angle and velocity when it reached the ground.

Answer 1

Given :

▪ Angle of projection = 30°

▪ Initial velocity = √5î + j

To Find :

▪ Angle and velocity when projectile reached the ground.

Concept :

↗ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a verticle plane such that its horizontal acceleration is zero and verticle downward acceleration is equal to g.

↗ In projectile motion, the horizontal motion and the verticle motion are independent of each other. i.e., neither motion affects the other.

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☆ The velocity at the end of flight of an oblique, projectile is the same in magnitude as at the beginning but the angle that it makes with the horizontal is negative at the angle of projection.

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$\implies\sf\:(angle)_f=-(angle)_i\\ \\ \implies\sf\:\theta_f=-(\theta)_i\\ \\ \implies\underline{\boxed{\bf{\red{\theta_f=-30\degree}}}}$

☢ Initial velocity :

$\mapsto\sf\:x-component=v_x\\ \\ \mapsto\sf\:v_x=v\cos\theta_i\\ \\ \mapsto\bf\:v\cos\theta_i=5\hat{i}\\ \\ \mapsto\sf\:y-component=v_y\\ \\ \mapsto\sf\:v_y=v\sin\theta_i\\ \\ \mapsto\bf\:v\sin\theta_i=\hat{j}$

☢ Final velocity :

$\implies\sf\:x-component=v_x\\ \\ \implies\sf\:v_x=v\cos\theta_f\\ \\ \implies\sf\:v\cos\theta_i=v\cos(-\theta)_f\\ \\ \implies\bf\:v_x=v\cos\theta_i=\sqrt{5}\hat{i}\\ \\ \implies\sf\:y-component=v_y\\ \\ \implies\sf\:v_y=v\sin\theta_f\\ \\ \implies\sf\:v\sin\theta_i=v\sin(-theta)_f\\ \\ \implies\sf\:v_y=-v\sin\theta_i=-\hat{j}\\ \\ \bigstar\:\underline{\boxed{\bf{\orange{Final\:velocity=\sqrt{5}\hat{i}-\hat{j}}}}}$

Answer 2

Answer

Given :

• Angle of projection = 30°
• Initial velocity = √5 i + j

Solution :

$\bold{(Angle)_f=(Angle)_i}\\\\\implies \bold{\theta_f=\theta_i}\\\\\implies \bold{\theta_f=-30^0}$

Initial Velocity :

x - component = $v_x$

$\implies \bold{v_x=vcos\theta_i}\\\\\implies \bold{v_x=\sqrt{5}i }$

y - component = $v_y$

$\implies \bold{v_y=vsin\theta_i}\\\\\implies \bold{v_y=j}$

Final Velocity :

x - component = $v_x$

$\implies \bold{v_x=vcos\theta_f}\\\\\implies \bold{v.cos\theta_i=v.cos(-\theta_f)}\\\\\implies \bold{v.cos\theata_i=v.cos\theta_f}\\\\\implies \bold{v_x=\sqrt{3}i }$

y - component = $v_y$

$\implies \bold{v_y=v.sin\theta_f}\\\\\implies \bold{v.sin\theta_i=v.sin(-\theta_f)}\\\\\implies \bold{v.sin\theta_f=-v.sin\theta_i}\\\\\implies \bold{v_y=-j}$

Hence the velocity when it reached the ground is √3 i - j .

Here angle will be same of angle of projection , but inverted . i.e., -ve.

Hence the angle = -Θ =  - 30°