Physics

Projectile Motion

36 Questions

Projectile motion involves the movement of an object thrown into the air, subject to gravity. These questions cover horizontal range, time of flight, and trajectory calculations. This topic is crucial for various competitive exams requiring a strong foundation in physics.

Time of flightHorizontal rangeAngle of projectionVertical projectionVelocity components

Projectile Motion Questions

Multiple choice physics universe and space science launching of an artificial satellites around the earth launching of satellite india’s space programmes

A particle is projected vertically upward and is at a height h after $\displaystyle { t } _{ 1 }$ seconds and again after $\displaystyle { t } _{ 2 }$ seconds, then:

  1. $\displaystyle h={ gt } _{ 1 }{ t } _{ 2 }$
  2. $\displaystyle h=\frac { 1 }{ 2 } { gt } _{ 1 }{ t } _{ 2 }$
  3. $\displaystyle h=\frac { 2 }{ g } { t } _{ 1 }{ t } _{ 2 }$
  4. $\displaystyle h=\sqrt { { gt } _{ 1 }{ t } _{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let u be the velocity of the projection and O be the point of projection. Let P be the point in the path of the particle such that OP=h. Then,
$\displaystyle h=ut-\frac { 1 }{ 2 } { gt }^{ 2 }\Rightarrow { gt }^{ 2 }-2ut+2h=0$...(i)
Clearly, $\displaystyle { t } _{ 1 }{ t } _{ 2 }$, are two roots of this equation.
$\displaystyle \therefore \quad { t } _{ 1 }{ t } _{ 2 }=\frac { 2h }{ g } $
$\displaystyle \Rightarrow \quad h=\frac { 1 }{ 2 } g{ t } _{ 1 }{ t } _{ 2 }$

Multiple choice physics universe and space science launching of an artificial satellites around the earth launching of satellite india’s space programmes

If for a given angle of projection, the horizontal range is doubled, the time of flight becomes

  1. $4 times$
  2. $2 times$
  3. $\sqrt 2 \;times$
  4. $\frac{1}{{\sqrt 2 }}\;times$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$r=\dfrac { { u }^{ 2 }\sin2\theta  }{ g } $
Hence $R=2r=\dfrac { { V }^{ 2 }2\sin2\theta  }{ g } $
${ u }^{ 2 }=\dfrac { { V }^{ 2 } }{ 2 } \Rightarrow V=\left[ { \left( 2 \right)  }^{ 1/2 } \right] 4$
$t=\dfrac { 2u\sin\theta  }{ g } $
$T=\dfrac { 2V\sin\theta  }{ g } $
    $=\dfrac { 2\times { 2 }^{ 1/2 }\times u\sin\theta  }{ g } $
    $=\left( { 2 }^{ 1/2 } \right) t$
    $={ \left( 2 \right)  }^{ 1/2 }$
    $=\sqrt { 2 } $ times
Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

A ball is projected from the ground at angle 0 with the horizontal. After 1 sec it is moving at angle ${ 45 }^{ \circ  }$ with the horizontal and after 2s it is moving horizontally. What is the velocity of projection of the ball ? (Take $g=10\quad { ms }^{ -2 }$)

  1. $10\sqrt { { 3ms }^{ -1 } } $
  2. $20\sqrt { { 3ms }^{ -1 } } $
  3. $10\sqrt { { 5ms }^{ -1 } } $
  4. $20\sqrt { { 2ms }^{ -1 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using kinematic equations for projectile motion: vx = v*cos(theta), vy = v*sin(theta) - gt. At t=1, vy/vx = tan(45) = 1. At t=2, vy = 0, so v*sin(theta) = 2g = 20. Solving these gives the initial velocity.

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

A stone is projected from the ground with a velocity of $14 \ ms^{-1}$ one second later it clean a wall $2 \ m$ high. The angle of projection is $(g = 10 \ ms^{-2})$ 

  1. $30 ^\circ$
  2. $45 ^\circ$
  3. $60 ^\circ$
  4. $15 ^\circ$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the equation of trajectory y = x*tan(theta) - (g*x^2)/(2*u^2*cos^2(theta)). With u=14, x = u*cos(theta)*t = 14*cos(theta)*1, and y=2, we solve for theta.

Multiple choice physics along with motion motion around us motion and rest moving things around us

If $\vec {u}=a \hat {i}+b \hat {j}+ c \hat {k}$ with $\hat {i},\hat {j},\hat {k}$ are in east, north and vertical directions, the maximum height of the projectile is ?

  1. $\dfrac{a^{2}}{2g}$
  2. $\dfrac{b^{2}}{2g}$
  3. $\dfrac{c^{2}}{2g}$
  4. $\dfrac{b^{2} c^{2}}{2g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum height of a projectile is H = (vertical component of velocity)²/(2g). If vector u = aî + bĵ + cĵ with î east, ĵ north (horizontal), and ĵ vertical, then the vertical component of initial velocity is c (assuming cĵ represents the vertical direction, which typically uses the unit vector k̂, but in this notation cĵ or ck̂ would give the vertical component c). Maximum height = c²/(2g).

Multiple choice physics motion and measurement of distances motion around us motion and rest moving things around us

The height $y$ and horizontal distance $x$ covered by a projectile in a time $t$ seconds are given by the equations $y = 8t - 5t^2$ and $x = 6t$. If $x$ and $y$ are measured in meters, the velocity of projection is:-

  1. $10 \ ms^{-1}$
  2. $6 \ ms^{-1}$
  3. $8 \ ms^{-1}$
  4. $14 \ ms^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$v _x=\dfrac{dx}{dt}=6$

$v _y=\dfrac{dy}{dt}=8-10t$
$\vec v=v _x\hat{i}+v _y\hat{j}$

$\vec u=\vec v _{t=0}$
$\vec u=6\hat{i}+8\hat{j}$
$u=\sqrt{6^2+8^2}$
$u=10m/s$

Multiple choice power work and power work, energy and power physics energy and its forms

A stone is projected with velocity $u$ at an angle $\theta$ with horizontal. Find out average power of the gravity during time t.

  1. $mg\, \left [ \displaystyle \frac{gt^2}{2}\, - u sin \theta \right ]$
  2. $mg\, \left [ \displaystyle \frac{gt}{2}\, + u sin \theta \right ]$
  3. $mg\, \left [ \displaystyle \frac{gt}{2}\, - u sin \theta \right ]$
  4. $mg\, \left [ \displaystyle \frac{gt}{4}\, - u sin \theta \right ]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average power of gravity = Work / time. Work done by gravity = -m * g * delta_y. Delta_y = u * sin(theta) * t - (1/2) * g * t^2. Work = -m * g * (u * sin(theta) * t - 0.5 * g * t^2). Power = Work / t = -m * g * (u * sin(theta) - 0.5 * g * t) = m * g * (0.5 * g * t - u * sin(theta)).

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

Two stones are projected with the same speed but making different angles with horizontal. Their ranges are equal. If the ranges of projection of one is $\pi/3$ and its maximum height is ${h} _{1}$ then the maximum height of the other will be

  1. $\dfrac{Rg}{\omega^{2}}$
  2. $\dfrac{R^{2}g}{\omega^{}}$
  3. $\dfrac{R^{2}g}{\omega^{2}}$
  4. $\dfrac{R^{2}\omega}{g^{}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Assuming this to be a geo-stationary satellite, the angular speed of the satellite is the same as the angular speed of rotation of the earth. 

Hence, we have:
$GMm/r^2=mw^2r$ and $GM/R^2=g$
Hence, $r^3=R^2g/w^2$

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

A particle is projected with a velocity $\sqrt{\dfrac{4gR}{3}}$ vertically upward from the surface of the earth. R is the radius of the earth & g being the acceleration due to gravity on the surface of the earth. The velocity of the particle when it is at half the maximum height reached by it is:

  1. $\sqrt{\dfrac{gR}{2}}$
  2. $\sqrt{\dfrac{gR}{3}}$
  3. $\sqrt{gR}$
  4. $\sqrt{\dfrac{2gR}{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using energy conservation: (1/2)mv^2 - GMm/R = -GMm/(R+h). Given v = sqrt(4gR/3) and g = GM/R^2, we find max height h. Then find velocity at h/2 using energy conservation again.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics
A cannonball is fired from a cannon at 50.0 m/s, at an angle of 30.0° above the horizontal on level ground. The goal is to determine the distance from its starting point to its ending point. The path of the cannonball is shown below. (inverse parabola).Calculate Δx.

  1. 15 cm

  2. 20 cm

  3. 30 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

In projectile motion, the horizontal component of velocity remains constant throughout the motion.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In projectile motion, the horizontal component of velocity remains constant because there is no acceleration in the horizontal direction.

Multiple choice

The range of a projectile is the maximum horizontal distance it travels before returning to the ground.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The range of a projectile depends on the initial velocity, launch angle, and the acceleration due to gravity.

Multiple choice

A projectile is fired from the ground with a velocity of 10 m/s at an angle of 30 degrees with the horizontal. What is the maximum height reached by the projectile?

  1. 5 m

  2. 10 m

  3. 15 m

  4. 20 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum height reached by the projectile is given by the equation $h = (u^2 sin^2 \theta)/(2g)$. Substituting the values, we get $h = (10^2 sin^2 30)/(2 * 9.8) = 10 m$.

Multiple choice

What is the primary factor that determines the range of a projectile?

  1. Initial velocity

  2. Mass of the projectile

  3. Air resistance

  4. Angle of projection
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle at which a projectile is launched significantly influences its range. A projectile launched at a higher angle will travel farther than one launched at a lower angle, assuming other factors remain constant.