Motion in a Plane · Physics · TS EAMCET
MCQ (Single Correct Answer)
The vertical displacement ( $y$ in metre) of a projectile in term of its horizontal displacement ( $x$ in metre) is given by $y=\left(\sqrt{3} x-0.2 x^2\right)$. The time of flight of the projectile is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
A body projected at certain angle $\left(\neq 90^{\circ}\right)$ from the ground crosses a point in its path at a time of 2.3 s and from there it reaches the ground after a time of 5.7 s . The maximum heigh reached by the body is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
A ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times 2 s and 8 s respectively. The horizontal distance between the two points is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
If a body projected with a velocity of $19.6 \mathrm{~ms}^{-1}$ reaches a maximum height of 9.8 m , then the range of the projectile is
(Neglect air resistance)
Two bodies are projected from the same point with the same initial velocity ' $u$ ' making angles ' $\theta^{\prime}$ and $\left(90^{\circ}-\theta\right)$ with the horizontal in opposite directions. The horizontal distance between their positions when the bodies are at their maximum heights is
A helicopter flying horizontally with a velocity of $288 \mathrm{~km} / \mathrm{h}$ drops a bomb. If the line joining the point of dropping the bomb and the point where bomb hits the ground makes an angle $45^{\circ}$ with the horizontal, then the height at which the bomb was dropped is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
The maximum horizontal range of a ball projected from the ground is 32 m . If the ball is thrown with the same speed horizontally from the top of a tower of . height 25 m , the maximum horizontal distance covered by the ball is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
A projectile is given an initial velocity of $\hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathrm{~ms}^{-1}$. The cartesian equation of its path is ( $x$ and $y$ are in metres and $g=10 \mathrm{~ms}^{-1}$ )
A player can throw a ball to a maximum horizontal distance of 80 m . If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is
The velocity of a particle having magnitude of $10 \mathrm{~ms}^{-1}$ in the direction of $60^{\circ}$ with positive $X$-axis is
A stone projected from the ground with a velocity $50 \mathrm{~ms}^{-1}$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of 3 s . Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
A person walks in such a way that he covers equal distance in each step. The person takes 2 steps forwards towards east, then takes a right turn and
walks 4 steps towards south, then takes a right turn and walks 6 steps towards west and then takes a right turn and walks further. The direction of his final position after a total of 20 steps walk with respect to his initial position is
Particle $A$ (which was located at the origin at time $t=0$ ) is moving along the $X$-axis with a constant speed of $1 \mathrm{~m} / \mathrm{s}$. Location of particle $B$ which is moving along the $Y$-axis is given by $y=c t^2$, where $c=1 \mathrm{~m} / \mathrm{s}^2$. Find the speed of particle $A$ relative to particle $B$ at $t=1 \mathrm{~s}$
A particle is moving in $X Y$-plane as $\mathbf{x}=\left(4 t+t^2\right) \hat{\mathbf{i}}$, $\mathbf{y}=\left(2 t+\frac{t^2}{2}\right) \hat{\mathbf{j}}$, where $\mathbf{x}$ and $\mathbf{y}$ are displacements measured along $X$ and $Y$-axes respectively, in metres and $t$ in seconds, What is the velocity of the particle?
The surface of a hill inclined at an angle $30^{\circ}$ to the horizontal. A stone is thrown from the summit of the hill (point $A$ ) at an initial speed $10 \mathrm{~m} / \mathrm{s}$ at angle $60^{\circ}$ to the vertical. If the stone strikes the hill at point $B$ as shown in the figure, the distance between $A$ and $B$ is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
Statement I An object subjected to velocities $\mathbf{v}_1$ and $\mathbf{v}_2$ has a resultant velocity with magnitude $|\mathbf{v}|=\left|\mathbf{v}_1\right|+\left|\mathbf{v}_2\right|$.
Statement II The magnitude of displacement is either less or equal to the path length of an object between two points.
Statement III The instantaeous acceleration is the limiting value of the average acceleration as the time interval approaches zero.
Which of the following is correct?
For a projectile, if $\alpha$ is the angle of projection, $R$ is the range, $h$ is the maximum height, $t$ is the time of flight then
Two cars, at a certain instant, are 50 km apart on a line running from south to north. The one farther north is moving west at $25 \mathrm{~km} / \mathrm{h}$. The other is moving towards north at $25 \mathrm{~km} / \mathrm{h}$. How long do they take to reach their distance of closest approach?
- A particle initially at origin starts moving in $X Y$ - plane has velocity component $\mathbf{v}=(6+2 t) \hat{\mathbf{i}}+(4+2 \sqrt{3 t}) \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$. Acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is $[x, y$ are measured in meters, $t$ in seconds, respectively
A bullet is fired at time $t=0$ with velocity $20 \mathrm{~m} / \mathrm{s}$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time 0.1 s is (assume $g=10 \mathrm{~m} / \mathrm{s}^2$ ).
A man walking along a straight line with a velocity 6 $\mathrm{km} / \mathrm{h}$ encounters rain falling vertically down with a velocity $6 \sqrt{3} \mathrm{~km} / \mathrm{h}$. At what angle the man should hold his umbrella, so that he can protect himself from rain
A projectile is given an initial velocity of $(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{\mathbf{j}}$ is along the vertical. Assuming $g=10 \mathrm{~m} / \mathrm{s}^2$, if the equation of its trajectory can be written as $\frac{1}{9}\left[\beta x+\gamma x^2\right]$. Then the value of $\gamma$ is
A small object slides down with initial velocity equal to zero from the top of a smooth hill of height $H$. The other end of the hill is horizontal and is at height $H / 2$ as shown in the figure. The horizontal distance covered by the object from the end of the hill to the ground is
A projectile is launched with an initial speed of $40 \mathrm{~m} / \mathrm{s}$ at an angle $30^{\circ}$ above the ground. The projectile lands on a hillside 2.0 s later. The net displacement from where the projectile lands on hillside 2.0 s later. The net displacement from where the projectile was launched to where it hits the target is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
A ball is projected with a velocity $5 \mathrm{~m} / \mathrm{s}$, so that its horizontal range is twice the greatest height attained. The value of range is
A point moves in the $x y$-plane according to the following equation, $x=a \sin \omega t, y=a(1-\cos \omega t)$, where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.
A particle of mass $m=1 \mathrm{~kg}$ moves in the $x y$-plane. The force on it at time $t$ is $F(t)=[2 \sin (\alpha t) \hat{\mathbf{i}}+3 \cos (\alpha t) \hat{\mathbf{j}}] \mathrm{N}$, where $\alpha=1 \mathrm{~s}^{-1}$. At time $t=0$, the particle is at rest at the origin. Calculate the magnitude of its position vector $\mathbf{r}$ (in m ) and velcoity vector $\mathbf{v}$ (in m/s) at time $t=\frac{\pi}{2} \mathrm{~s}$.
A particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by 10 m . If projected with an angle of $45^{\circ}$ is away from the target by 10 m , then the angle of projection to hit the target is
A person moves 30 m North and then 20 m towards East and finally $30 \sqrt{2} \mathrm{~m}$ in South-West direction. The displacement of the person from the origin will be
A river 200 m wide is flowing at a rate of $3.0 \mathrm{~m} / \mathrm{s}$. A boat is sailing at a velocity of $15 \mathrm{~m} / \mathrm{s}$ with respect to the water in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach on the opposite bank?
Two cars $A$ and $B$ are moving with speeds $v_A=120 \mathrm{km} / \mathrm{h}$ and $v_B=50 \mathrm{~km} / \mathrm{h}$ respectively in the directions as indicated by the arrow in the figure below. What is the relative speed of the car $B$ with respect to car $A$ ?
Initial velocity with which a body is projected is $10 \mathrm{~m} / \mathrm{s}$ from the base of an inclined plane as shown in the given figure. If the angle of projection is $60^{\circ}$ with the horizontal, then the range $R$ is [take, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
A projectile is launched from point $A$ of the given landscape with a water body as shown in the diagram. The launching angle is $15^{\circ}$. From the following, identify the right initial velocity of the projectile with which it will fall somewhere in between the points $C$ and $D$. [Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
