The force of attraction of the earth on anybody is called the force of gravity. Acceleration produced on the body by the force of gravity is called acceleration due to gravity. It is represented by the symbol ‘g’.
Sign convention:
There are three cases basically in this -
1) If a body dropped from some height (initial velocity zero)
$
\begin{aligned}
& \mathrm{u}=0 \\
& \mathrm{a}=\mathrm{g} \\
& v=g t \\
& h=\frac{1}{2} g t^2 \\
& v^2=2 g h \\
& h_n=\frac{g}{2}(2 n-1)
\end{aligned}
$
2) If a body is projected vertically downward with some initial velocity
$
\begin{aligned}
& \text { Equation of motion: } \quad v=u+g t \\
& \qquad \begin{array}{l}
h=u t+\frac{1}{2} g t^2 \\
v^2=u^2+2 g h \\
h_n=u+\frac{g}{2}(2 n-1)
\end{array}
\end{aligned}
$
3) If a body is projected vertically upward.
(i) Apply equation of motion :
Take initial position as origin and the direction of motion (vertically up) as $\mathrm{a}=-\mathrm{g} \quad$ [as acceleration due to gravity is downwards]
So, if the body is projected with velocity $u$, and after time t it reaches up to height h then,
$
\mathrm{v}=\mathrm{u}-\mathrm{g} t ; \quad \mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^2 ; \mathrm{v}^2=\mathrm{u}^2-2 \mathrm{gh}
$
(ii) For the case of maximum height $v=0$
So from the above equation
$
\begin{gathered}
u=g t \\
h=\frac{1}{2} g t^2 \\
\text { and } u^2=2 g h
\end{gathered}
$
| Exam | Chapter |
| JEE MAIN | Kinematics |
A balloon is moving up in the air vertically above point A on the ground. When it is at a height , a girl standing at a distance (point
) from
(see figure) sees it at an angle
with respect to the vertical. When the balloon climbs up a further height
, it is seen at an angle
with respect to the vertical if the girl moves further by a distance
(point
). Then the height
is (given
) :

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A helicopter rises from rest on the ground vertically upwards with a constant acceleration . A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to [
is the acceleration due to gravity ]:
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A tennis ball is released from a height $h$ and after freely falling on a wooden floor, it rebounds and reaches height $\frac{h}{2}$. The velocity versus height of the ball during its motion may be represented graphically by:
(graphs are drawn schematically and not to the scale)
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Distance travelled (in meters) by a free falling body during 3rd second of its motion is ( g = 10m/s2)
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A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s2. He reaches the ground at a speed of 3 m/s. At what height (in meters), did he bail out?
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A ball of mass 0.5 kg is dropped from a height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is $\qquad$ m _______. [Use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
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A ball is projected vertically upward with an initial velocity of ,another ball is projected vertically upward with the same
______ s, the second ball will meet the first ball
.
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When a ball is dropped to a lake from a height 4.9 m above the water level, it hits the water with a velocity $v$ and then sinks to the bottom with the constant velocity $v$. It reaches the bottom of the lake 4.0 after it is dropped. The approximate depth of the lake is :
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Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at $\mathrm{t}=0 \mathrm{~s}$. Ball B is thrown vertically down with an initial velocity ${ }^{\prime} \mathrm{u}^{\prime}$ at $\mathrm{t}=2 \mathrm{~s}$. After a certain time, both balls meet 100 m above the ground. Find the value of ' u ' in $\mathrm{ms}^{-1}$. [use $g=10 \mathrm{~ms}^{-2}$ ] :
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A NCC parade is going at a uniform speed of $9 \mathrm{~km} / \mathrm{h}$ under a mango tree on which a monkey is sitting at a height of 19.6 m . At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ )
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A ball is thrown vertically upwards with a velocity of $19.6 \mathrm{~ms}^{-1}$ from the top of a tower. The ball strikes the ground after $6s$. The height from the ground up to which the ball can rise will be $\left(\frac{k}{5}\right) \mathrm{m}$. The value of $k$ is___________ (use $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ )
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A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is
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A ball is released from a height $h$. If $t_1$ and $t_2$ are the time required to complete the first half and second half of the distance respectively. Then, choose the correct relation between $t_1$ and $t_2$.
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Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{\text {th }}$ the second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take
$
\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}_{\text {) }}^2
$
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A balloon was moving upwards with a uniform velocity of $10 m/s$. An object of finite mass is dropped from the balloon when it was at a height of $75 m$ from the ground level. The height of the balloon from the ground when object strikes the ground was around:
(takes the value of $g$ as $10 \mathrm{~m} / \mathrm{s}^2$ )
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Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at regular intervals of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of the second drop from the floor when the first drop strikes the floor.
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A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and the other downwards) with the same velocity u and they reach the earth's surface after and
seconds respectively, then
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From a building two balls and
are thrown such that
is thrown upwards and
downwards (both vertically). If
and
are their respective velocities on reaching the ground, then
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A ball is released from the top of a tower of height metre. It takes
second to reach the ground. What is the position of the ball in
second?
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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out?
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A ball is dropped from the top of a building 100m high. At the same time, another ball is thrown upwards with a velocity of 40 m/sec from the bottom of the building. The two balls will meet after T seconds. The value of 10T will be -
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From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in _____s.
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From a tower of height H, a particle is thrown vertically upwards with a speed of u. The time taken by the particle, to hit the ground is n times that taken by it to reach the highest point of its path.
The relation between H, u and n are:
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A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is ____ m.
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A body falling from rest under gravity passes a certain point $P$. It was at a distance of 400 m from $\mathrm{P}, 4 \mathrm{~s}$ before passing through $P$. If $g=10^2 \mathrm{~m} / \mathrm{s}^2$, then the height above the point P from where the body began to fall is:
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By what velocity (in m/s) should a ball be projected vertically downward so that the distance covered by it in the 5th second is twice the distance it covers in its 6th second? ( g= 10 m / s2 )
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The magnetic susceptibility of a material of a rod is 499 . Permeability in vacuum is . Absolute permeability of the material of the rod is :
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A skydiver drops 50 metres without friction after a fall. When the parachute opens, it decelerates at It reaches the ground at a speed of
How high was he released on bail?
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If $S_1=$ displacement in $4^{\text {th }}$ second for freely falling body
$\mathrm{S}_2=$ Displacement of freely body falling body in 4 sec.
Then $\frac{S_1}{S_2}$ will be:
$\left(\right.$ take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
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A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?
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A ball dropped from the 16th storey of a multi-storeyed building reaches the ground in 4 seconds. In the 3rd second of its free fall, it passes through n storeys, where n is equal to (Take g =10 ms-2)
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A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height $h$.
Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.
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A ball is thrown up with a certain velocity so that it reaches a height $h$ Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.
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A ball is thrown vertically upward with a velocity of 20 ms-1 from a top of a multistory building. The height of the point from where the ball is thrown is 25 m from the ground. The height to which the ball rises from the ground and time taken before the ball hits the ground is (g:=10 ms-2)
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A ball is thrown vertically upward with an initial velocity of $150 \mathrm{~m} / \mathrm{s}$. The ratio of velocity after $3 s$ and $5 s$ is $\frac{x+1}{x}$. The value of $x$ is _____________.
$\left\{\right.$ take,$\left.g=10 \mathrm{~m} / \mathrm{s}^2\right\}$
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A stone is dropped from a height h, simultaneously another stone is thrown up from the ground which reaches the maximum height 3h, the two stones cross each other after time
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A tennis ball is dropped on to the floor from a height of 9.8 m .It rebounds to a height 5.0 m . Ball comes in contact with the floor for 0.2 s . The average acceleration during contact is $\mathrm{ms}^{-2}$
(Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
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A 30.0 kg mass falls from a height of 4.0 m. The momentum of the mass just before it hits the ground is
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A body falling under gravity covers two points A and B separated by 80 m is 2s. The distance of upper point A from the starting point is _____ m (use g = 10 ms–2)
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A body is thrown vertically upward with velocity u. The distance travelled by it in the 5th and 6th seconds is equal. The displacement in 5th seconds is equal to (take g=10 m/s2 )
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An object is released from the top of a building. Its a-t graph is:
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A rescue chopper is rising vertically up with a speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$. A bomb is thrown in an upward direction from the chopper with a speed $V_1$ with respect to the chopper. The bomb crosses the chopper 3 seconds after it was thrown. What is the value of $V_1$ in m/s?
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Two balls A and B are thrown from the top of the tower one vertically upward and the other vertically downward with the same speed. If times taken by them to reach the ground are 4 s and 2 s respectively, then the height of the tower and initial speed of each ball are (g=10 m/s2)
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A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a fraction r of the speed with which it strikes the latter on each impact, what is the net distance travelled by the ball up to the 10th impact?
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A boy is standing on top of a tower of height 85 m and throws a ball in the vertically upward direction with a certain speed. If 5.25 seconds later he hears the ball hitting the ground, then the speed with which the boy threw the ball is (take g = 10 m/s2 , speed of sound in air = 340 m/s)
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A student performs an experiment to determine the acceleration due to gravity g. The student throws a steel ball up with initial velocity u and measures the height h traveled by it at different times t. The graph the student should plot on graph paper to readily obtain the value of g is
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If $S_1=$ Displacement in $4^{\text {th }}$ second for a freely falling body and $S_2=$ Displacement of a freely falling body in 4 sec. Then find $\left(\frac{S_1}{S_2}\right) \quad$ (take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$ is x/16 then the value of x is
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Two balls A and B are thrown from the ground, A is thrown up and B is thrown down (Both vertically). If vA and vB are their respective speeds when they hit the ground, then choose the correct option.
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