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Projectile Motion - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Projectile Motion is considered one the most difficult concept.

  • 56 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

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

The relation between the time of flight of projectile {{T}_{f}} and the time to reach the maximum height {{t}_{m}} is

Assertion: In the plane-to-plane motion of a projectile, the horizontal component of velocity remains constant.

Reason: In plane-to-plane projectile motion, the horizontal component of acceleration is zero.

A projectile fired with initial velocity  v at an angle \theta. If initial velocity be tripled at the same angle of projection, then time of ascent will become 

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Concepts Covered - 1

Projectile Motion
  • Projectile Motion- Two dimesional motion under action of uniform force is called is called projectile motion.

                E.g-  A javelin thrown by an athlete 

1. Projectile Projected at an angle \theta

  • 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 - 

  1. Maximum Height - 

 

  • Maximum vertical distance attained by a projectile during its journey.

  • Formula-H= \frac{U^{2}\sin ^{2}\theta }{2g}

  • When the velocity of projectile increases n time then Maximum height is increased by a factor of  n^2

  • Special Case-

                                If U is doubled, H becomes four times provided \theta & g are constant.

 

     2)  Time of Flight

  • Time for which projectile remains in the air above the horizontal plane.

  • Formula-

  1.  T = \frac{2usin\theta}{g}

  2. Time of ascent =

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

 

  1. 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 the projectile increased n time then the time of flight becomes n times.

 

  1. Horizontal Range

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

  • Formula-

                         R=\frac{u^{2}\sin 2\theta }{g}

  • Special case of horizontal range

  1. 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}

 

  1. Range remains the same whether the projectile is thrown at an angle \theta with the horizontal or at an angle \theta with vertical (90-\theta) with horizontal 

  2. When the velocity of projectile increases n time then the horizontal range is increased by a factor of  n^{2}

  3. When the horizontal range is n times the maximum height then 

                                                                                   \tan \theta= \frac{4}{n}

  • Equation of trajectory-
  • y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}

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Projectile Motion

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