Summary for Time, Height and Range

Q.

The horizontal range of an oblique projectile is four times the maximum height attained by the projectile. The angle of projection is

1. $45°$

2. $30°$

3. $60°$

4. $90°$

Q.

Two bodies are projected with the same velocity. If one is projected at an angle of $30°$ and the other at an angle of $60°$ to the horizontal, find the ratio of the maximum heights reached.

1. $\frac{1}{4}$

2. $\frac{1}{3}$

3. $\frac{1}{5}$

4. $\frac{1}{9}$

Q.

A stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following

1. Straight path

2. Circular path

3. Parabolic path

4. Hyperbolic path

Q.

A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of ${30}^{\circ }$ with the horizontal. The change in momentum (in magnitude) of the body is

1. 24.5 N-s

2. 49.0 N-s

3. 98.0 N-s

4. 50.0 N-s

Q.

A stone is projected from the ground with velocity 25 m/s. Two seconds later, it just clears a wall 5 m high. The angle of projection of the stone is $(g=10m/se{c}^2)$

1. ${30}^{\circ }$

2. ${45}^{\circ }$

3. ${50.2}^{\circ }$

4. ${60}^{\circ }$

Q. A boy throws a water-filled balloon at an angle of ${53}^{\circ }$ with a speed of $10m/s$. A car is advancing towards the boy at a constant speed of $5m/s$. If the balloon is to hit the car, how far away should the car be when the balloon is thrown? $(g=10m{s}^{-2})$

1. $8m$
2. $9.6m$
3. $15.6m$
4. $17.6m$

Q.

A particle of mass m is projected with velocity v making an angle of ${45}^{\circ }$ with the horizontal. Themagnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

1. $Zero$

2. $\frac{m{v}^3}{\left(4\sqrt{2}g\right)}$

3. $\frac{m{v}^3}{\left(\sqrt{2}g\right)}$

4. $\frac{m{v}^2}{2g}$

Q. A projectile is thrown with a velocity of $50m{s}^{-1}$ at an angle of ${53}^o$ with the horizontal. The equation of the trajectory is given by (Take $g=10m/s^2$)

1. $180y=240x-x^2$
2. $180y=x^2-240x$
3. $180y=135x-x^2$
4. $180y=x^2-135x$

Q. Trajectory of two particles projected from origin with speeds $v_1$ and $v_2$ at angles ${\theta }_1$ and ${\theta }_2$ with positive x-axis respectively are as shown in the figure. Given that $g=-10{\text{m/s}}^2\left(\hat{j}\right)$. Choose the correct option(s) related to diagram.

1. $v_1-v_2=2v_1$
2. ${\theta }_2-{\theta }_1=2{\theta }_1$
3. $v_1=v_2$
4. $3({\theta }_2-{\theta }_1)=+{\theta }_1$

Q. When a body is projected with velocity $\overrightarrow{v},$ horizontally from a height $h$.

$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(a) The speed of body at any time}t\text{is}& \text{(p)}\sqrt{\frac{2gh}{u^2}}\\ \text{(b) It will strike the ground after time}t\text{is equal to}& \text{(q)}\sqrt{\frac{2h}{g}}\\ \text{(c) The path followed by the body is expressed as}y\text{equal to}& \text{(r)}\frac{1}{2}g\frac{x^2}{u^2}\\ \text{(d) The value of}\tan \theta \text{, where}\theta \text{is the angle made by velocity striking the ground with horizontal is equal to}& \text{(s)}\sqrt{v^2+g^2{t}^2}\end{array}$

1. $a-r,b-p,c-q,d-s$
2. $a-r,b-q,c-p,d-s$
3. $a-r,b-p,c-r,d-s$
4. $a-s,b-q,c-r,d-p$

Q. A particle is projected with velocity $u$ at angle ${\theta }_1$ with horizontal. The ratio of range and maximum height is $4$.
When it is projected at angle ${\theta }_2$ with horizontal with same speed, the ratio of range and maximum height is $2$.
Then $\frac{\tan {\theta }_1}{\tan {\theta }_2}$ will be equal to

1. $4$
2. $2$
3. $\frac{1}{2}$
4. $\frac{1}{4}$

Q. If the horizontal range is four times the maximum height attained by a projectile then the angle of projection is

1. ${90}^{\circ }$
2. ${60}^{\circ }$
3. ${30}^{\circ }$
4. ${45}^{\circ }$

Q. The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

1. ${90}^{\circ }$
2. ${60}^{\circ }$
3. ${45}^{\circ }$
4. ${30}^{\circ }$

Q.

A stone projected with a velocity u at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity u at an angle $(\frac{\pi }{2}-\theta )$ with the horizontal, it reaches maximum height $H_2$.
The relation between the horizontal range R of the projectile, $H_1$ and $H_2$ is

1. $R=4\sqrt{H_1{H}_2}$

2. $R=4(H_1-H_2)$

3. $R=4(H_1+H_2)$

4. $R=\frac{H_1^2}{H_2^2}$