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Projectile Motion MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

Projectile Motion is considered one the most difficult concept.
58 Questions around this concept.

Solve by difficulty

A boy can throw a stone up to a maximum height of 10m . The maximum horizontal distance that the boy can throw the same stone up to will be :

$20 \sqrt{2}m$

$10m$

$10 \sqrt{2}m$

$20m$

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A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$ , the total area around the fountain that gets wet is

$\pi \frac{v^{2}}{g}$

$\pi \frac{v^{4}}{g^{2}}$

$\frac{\pi}{2} \frac{v^{4}}{g^{2}}$

$\pi \frac{v^{2}}{g^{2}}$

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Concepts Covered - 1

Projectile Motion

Projectile Motion-An object which is thrown with initial velocity & then it moves due to the effect of gravity alone is called projectile , it motion is called as projectile motion and its path called trajectory.

E.g- A javelin thrown by an athlete

1. Projectile Projected at angle ϴ

Initial Velocity- U

Horizontal component = $U_{x} = Ucos\theta$

Vertical component = $U_{y} = Usin\theta$

Final velocity = V

Horizontal component = $V_{x} = Ucos\theta$

Vertical component = $V_{y} = Usin\theta - g.t$

So,

$V = \sqrt{V^{2}_{x} +V^{2}_{y}}$

Displacement = S

Horizontal component = $S_{x} = Ucos\theta.t$

Vertical component = $S_{y} = Usin\theta.t - \frac{1}{2}.g.t^{2}$

and, $S = \sqrt{S^{2}_x + S^{2}_y}$

Acceleration = a

Horizontal component = 0

Vertical component = -g

So, a = -g

Parameters in Projectile motion -

Maximum Height -

Maximum vertical distance attained by projectile during its journey.
Formula- $H= \frac{U^{2}\sin ^{2}\Theta }{2g}$
When the velocity of projectile increased n time then Maximum height is increased by a factor of
Special Case-

If U is doubled , H become four times provided & g are constant.

2) Time of Flight

Time for which projectile remain in the air above the horizontal plane.
Formula-

$T = \frac{2Usin\theta}{g}$
Time of ascent =

$t_{a} = \frac{T}{2}$

Time of descent =

$t_{d} = \frac{T}{2}$

When the velocity of projectile increased n time then Time of ascent becomes n times
When the velocity of projectile increased n time then Time of descent becomes n times
When the velocity of projectile increased n time then time of flight becomes n times.

Horizontal Range

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.
Formula-

$R=\frac{u^{2}\sin 2\Theta }{g}$

Special case of horizontal range

For max horizontal range.

$\Theta = 45^{0}$

$R_{max}=\frac{u^{2}\sin 2 (45) }{g}$ $=\frac{u^{2}\times 1}{g}$ $=\frac{u^{2}}{g}$

Range remain the same whether the projectile is thrown at angle $\Theta$ with the horizontal or at angle $\Theta$ with vertical (90- $\Theta$ ) with horizontal
When the velocity of projectile increased n time then horizontal range is increased by a factor of $n^{2}$
When horizontal range is n times the maximum height then