Motion under Gravity

1. A canon shoots a ball upwards with an initial speed of 100 m/s. The to

A canon shoots a ball upwards with an initial speed of 100 m/s. The total time of flight of the ball is 20 s before it hits the ground. The ball looses 70% of its speed after hitting the ground. Which among the following is the correct height that the ball will bounce up after its first bounce? (g=10 m/s²)

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100 m
70 m
50 m
45 m
This question was previously asked in
UPSC CAPF – 2024
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The ball is shot upwards with an initial speed of 100 m/s. In vertical motion under gravity (g=10 m/s²), the time taken to reach the maximum height is given by t_up = u/g = 100/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% – 70%) of 100 m/s = 30% of 100 m/s = 0.30 * 100 m/s = 30 m/s (upwards). To find the height the ball bounces up, we use the equation v² = u² + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m/s, and the acceleration is a=-g=-10 m/s². So, 0² = 30² + 2 * (-10) * h. This simplifies to 0 = 900 – 20h, so 20h = 900, which gives h = 900/20 = 45 meters.
Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.
The collision with the ground is inelastic as the ball loses speed. The coefficient of restitution (e) for this bounce would be the ratio of the speed after bounce to the speed before bounce, i.e., e = 30/100 = 0.3. The maximum height reached after a bounce with speed v is given by h = v² / (2g).

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2. Which one of the following holds true for a freely falling object ?

Which one of the following holds true for a freely falling object ?

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It moves with a uniform velocity.
It moves with a uniform speed.
It moves with a non-uniform acceleration.
It moves with a uniform acceleration.
This question was previously asked in
UPSC CAPF – 2023
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The correct answer is D. A freely falling object near the Earth’s surface, neglecting air resistance, moves with a constant acceleration.
– Free fall is defined as motion under the sole influence of gravity.
– Near the Earth’s surface, the acceleration due to gravity ($\text{g}$) is approximately constant (about 9.8 m/s²).
– This constant acceleration causes the velocity of the object to change uniformly over time.
In reality, air resistance affects falling objects, causing their acceleration to decrease as their speed increases, eventually reaching a terminal velocity where acceleration is zero. However, in standard physics problems where “freely falling” is used, air resistance is usually neglected, assuming movement solely under gravity with uniform acceleration.

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3. Which one of the following statements for an object falling freely und

Which one of the following statements for an object falling freely under the influence of gravity is correct?

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Zero acceleration always implies zero velocity
Zero acceleration has no relation with the velocity of the object
Zero velocity at any instant necessarily means zero acceleration at that instant
Acceleration is constant all throughout the free fall
This question was previously asked in
UPSC CAPF – 2020
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Acceleration is constant all throughout the free fall.
An object falling freely under the influence of gravity experiences a constant acceleration equal to the acceleration due to gravity (g), assuming air resistance is negligible. This acceleration acts downwards.
Zero acceleration implies constant velocity, not necessarily zero velocity (A is incorrect). Zero acceleration means there is no net force acting on the object; velocity could be constant and non-zero (B is incorrect). An object thrown upwards momentarily has zero velocity at its peak, but the acceleration due to gravity is still acting on it at that instant (C is incorrect).

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4. Which one of the following statements is correct ?

Which one of the following statements is correct ?

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Acceleration due to gravity decreases with the increase of altitude
Acceleration due to gravity increases with the increase of depth (assuming earth to be a sphere of uniform density)
Acceleration due to gravity decreases with the increase of latitude
Acceleration due to gravity is independent of the mass of the earth
This question was previously asked in
UPSC CAPF – 2016
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Statement A is correct. The acceleration due to gravity (g) on the surface of the Earth is given by the formula g = GM/R², where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth. As altitude increases, R increases. Since g is inversely proportional to R², an increase in R leads to a decrease in g.
Understanding how the acceleration due to gravity varies with altitude, depth, latitude, and the mass of the celestial body is essential.
Statement B is incorrect: Acceleration due to gravity decreases with increasing depth inside the Earth. At depth d below the surface, g’ = g(1 – d/R), where R is the radius of the Earth. It becomes zero at the Earth’s center.
Statement C is incorrect: Acceleration due to gravity increases with increasing latitude, from the equator to the poles. This is due to the Earth’s equatorial bulge (closer to the center at poles) and the effect of Earth’s rotation (centrifugal force is maximum at the equator).
Statement D is incorrect: Acceleration due to gravity is directly proportional to the mass of the Earth (M).

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5. The acceleration due to gravity on the surface of the Earth is maximum

The acceleration due to gravity on the surface of the Earth is maximum and it

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increases as we go up
decreases as we go up or down
increases as we go down
neither increases nor decreases as we go up or down
This question was previously asked in
UPSC CAPF – 2011
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The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down.
The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM/r². On the surface, r is the Earth’s radius (R), so g_surface = GM/R². As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r/r², where M_r is the mass within radius r). Gravity is zero at the Earth’s center.
Therefore, gravity is maximum at the surface and decreases both above and below the surface. Minor variations exist due to altitude, latitude (Earth’s bulge), and local geological variations, but the general trend is decrease away from the surface.

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6. During free fall of an object :

During free fall of an object :

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its potential energy increases and its kinetic energy decreases.
its potential energy decreases and its kinetic energy increases.
both its potential energy and kinetic energy increase.
both its potential energy and kinetic energy decrease.
This question was previously asked in
UPSC CAPF – 2010
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During free fall of an object, its potential energy decreases and its kinetic energy increases.
– Free fall is defined as motion under the influence of gravity only.
– Potential energy (PE) due to gravity is given by mgh, where m is mass, g is acceleration due to gravity, and h is height. As an object falls, its height (h) decreases, so its potential energy decreases.
– Kinetic energy (KE) is given by 0.5mv², where m is mass and v is velocity. As an object falls under gravity, its speed (v) increases (due to acceleration ‘g’), so its kinetic energy increases.
– In ideal free fall (neglecting air resistance), the total mechanical energy (PE + KE) is conserved. The decrease in potential energy is equal to the increase in kinetic energy.
If air resistance is considered, some of the potential energy is converted into heat and sound energy due to air friction, and the increase in kinetic energy is less than the decrease in potential energy. However, the question refers to ‘free fall’ which typically implies ideal conditions unless specified otherwise.

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7. There is a ball of mass 320 g. It has 625 J potential energy when rele

There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is

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62·5 m/s
2·0 m/s
50 m/s
40 m/s
This question was previously asked in
UPSC NDA-2 – 2024
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The correct answer is A) 62·5 m/s.
When the ball is released freely from a height, its potential energy is converted into kinetic energy as it falls. By the principle of conservation of mechanical energy (assuming no air resistance), the potential energy at the initial height is equal to the kinetic energy just before hitting the ground.
The mass of the ball is m = 320 g = 0.320 kg.
The potential energy at the initial height is PE = 625 J.
Assuming the ball starts from rest, its initial kinetic energy is 0. When it hits the ground, its potential energy becomes 0 (taking the ground as the reference level).
By conservation of energy, Initial Total Energy = Final Total Energy.
PE_initial + KE_initial = PE_final + KE_final
625 J + 0 J = 0 J + KE_final
So, KE_final = 625 J.
The kinetic energy is given by KE = (1/2) * m * v², where v is the speed.
625 = (1/2) * 0.320 * v²
625 = 0.160 * v²
v² = 625 / 0.160 = 625000 / 160 = 62500 / 16
v = sqrt(62500 / 16) = sqrt(62500) / sqrt(16) = 250 / 4 = 125 / 2 = 62.5 m/s.

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8. The motion of a particle of mass m is described by the relation, $y =

The motion of a particle of mass m is described by the relation, $y = ut – \frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is

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$F = mleft( rac{du}{dt} ight)$
$F = mg$
$F = mleft( rac{dy}{dt} ight)$
$F = -mg$
This question was previously asked in
UPSC NDA-2 – 2023
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The force acting on the particle is $F = -mg$.
The given equation of motion, $y = ut – \frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\frac{dy}{dt} = u – gt$. The second derivative gives acceleration: $\frac{d^2y}{dt^2} = -g$. According to Newton’s second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.
The equation $y = ut – \frac{1}{2}gt^2$ is the standard kinematic equation for vertical displacement under constant gravitational acceleration ($g$), with initial velocity $u$. The force responsible for this motion is the gravitational force, which is $mg$ acting downwards. Assuming the upward direction as positive y, the downward force is represented as $-mg$.

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9. A tennis ball is thrown in the vertically upward direction and the bal

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

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8 m/s
12 m/s
16 m/s
20 m/s
This question was previously asked in
UPSC NDA-2 – 2021
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The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m).
At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g \approx 10 \, \text{m/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) \implies 0 = u^2 – 400 \implies u^2 = 400 \implies u = 20 \, \text{m/s}$. Using $g \approx 9.8 \, \text{m/s}^2$: $0^2 = u^2 + 2(-9.8)(20) \implies 0 = u^2 – 392 \implies u^2 = 392 \implies u \approx 19.8 \, \text{m/s}$. Both approximations are closest to 20 m/s among the given options.
This is a standard problem in projectile motion under constant acceleration (gravity). The time taken to reach the maximum height can be found using $v = u + at$. The total time of flight is twice the time to reach the maximum height (assuming it lands at the same level).

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10. A rigid body of mass 2 kg is dropped from a stationary balloon kept at

A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9.8 m/s²)

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980 m.s⁻¹ and 980 J
√980 m.s⁻¹ and √980 J
980 m.s⁻¹ and √980 J
√980 m.s⁻¹ and 980 J
This question was previously asked in
UPSC NDA-2 – 2019
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The correct option is D) √980 m.s⁻¹ and 980 J.
The problem involves calculating the final speed of a falling object using kinematics and determining the total mechanical energy at the beginning of the fall using the principle of conservation of energy.
Given: mass (m) = 2 kg, initial height (h) = 50 m, initial velocity (u) = 0 m/s, acceleration due to gravity (g) = 9.8 m/s².
1. Speed when it just touches the ground (v): Using the kinematic equation v² = u² + 2gh, we get v² = 0² + 2 * 9.8 * 50 = 980. So, v = √980 m/s.
2. Total energy when dropped: At the moment of dropping from a height h with zero initial velocity, the total mechanical energy is the sum of potential energy and kinetic energy. Potential Energy (PE) = mgh = 2 kg * 9.8 m/s² * 50 m = 980 J. Kinetic Energy (KE) = ½mu² = ½ * 2 kg * (0 m/s)² = 0 J. Total Energy = PE + KE = 980 J + 0 J = 980 J. (Note: By conservation of energy, the total energy remains 980 J just before hitting the ground, where all potential energy is converted to kinetic energy).

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