The trajectory of a projectile in a vertical plane is $y = \alpha x - \beta x ^ { 2 }$, where $\alpha$ and $\beta$ are constants and $x \& y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained $H$ are respectively given by
(1) $\tan ^ { - 1 } \alpha , \frac { 4 \alpha ^ { 2 } } { \beta }$
(2) $\tan ^ { - 1 } \left( \frac { \beta } { \alpha } \right) , \frac { \alpha ^ { 2 } } { \beta }$
(3) $\tan ^ { - 1 } \beta , \frac { \alpha ^ { 2 } } { 2 \beta }$
(4) $\tan ^ { - 1 } \alpha , \frac { \alpha ^ { 2 } } { 4 \beta }$

A bomb is dropped by a fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a:
(1) straight line vertically down the plane
(2) parabola in a direction opposite to the motion of plane
(3) parabola in the direction of motion of plane
(4) hyperbola

Two projectiles are thrown with same initial velocity making an angle of $45^{\circ}$ and $30^{\circ}$ with the horizontal respectively. The ratio of their respective ranges will be
(1) $1 : \sqrt { 2 }$
(2) $\sqrt { 2 } : 1$
(3) $2 : \sqrt { 3 }$
(4) $\sqrt { 3 } : 2$

A projectile is projected with velocity of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\theta$ will be : [use $\mathrm { g } = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ ]
(1) $\frac { 1 } { 2 } \sin ^ { - 1 } \left( \frac { 5 t ^ { 2 } } { 4 R } \right)$
(2) $\frac { 1 } { 2 } \sin ^ { - 1 } \left( \frac { 4 R } { 5 t ^ { 2 } } \right)$
(3) $\tan ^ { - 1 } \left( \frac { 4 t ^ { 2 } } { 5 R } \right)$
(4) $\cot ^ { - 1 } \left( \frac { R } { 20 t ^ { 2 } } \right)$

A ball is projected from the ground with a speed $15 \mathrm{~m}\mathrm{~s}^{-1}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal, then $\tan\theta$ will be equal to
(1) $\frac{1}{4}$
(2) $\frac{1}{2}$
(3) 2
(4) 4

A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?
(1) 25 m
(2) 50 m
(3) 100 m
(4) 200 m

A ball is projected with kinetic energy $E$, at an angle of $60 ^ { \circ }$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
(1) Zero
(2) $\frac { E } { 2 }$
(3) $\frac { E } { 4 }$
(4) $E$

An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be $\_\_\_\_$ degree.

A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{\circ}$, the ratio of instantaneous velocity to its average velocity is $\pi : x\sqrt{2}$. The value of $x$ will be
(1) 2
(2) 5
(3) 1
(4) 7

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$. Assertion A: When a body is projected at an angle $45^{\circ}$, its range is maximum. Reason R: For maximum range, the value of $\sin 2\theta$ should be equal to one. In the light of the above statements, choose the correct answer from the options given below:
(1) $A$ is false but $R$ is true
(2) $A$ is true but $R$ is false
(3) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$
(4) Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

A projectile is projected at $30 ^ { \circ }$ from horizontal with initial velocity $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The velocity of the projectile at $\mathrm { t } = 2 \mathrm {~s}$ from the start will be:
(1) $40 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) Zero
(3) $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $20 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$

A projectile fired at $30^\circ$ to the ground is observed to be at same height at time 3 s and 5 s after projection, during its flight. The speed of projection of the projectile is $\_\_\_\_$ $\mathrm{m}\mathrm{~s}^{-1}$. (Given $g = 10$ m s$^{-2}$)

A body of mass $m$ is projected with a speed $u$ making an angle of $45^\circ$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\dfrac{\sqrt{Z}\, m u^3}{X g}$. The value of $X$ is $\_\_\_\_$.

Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is:
(1) $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
(2) $\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$
(3) $\frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$
(4) $\frac{1 + \sin\alpha}{1 - \sin\alpha}$

The maximum speed of a boat in still water is $27\,\mathrm{km/h}$. Now this boat is moving downstream in a river flowing at $9\,\mathrm{km/h}$. A man in the boat throws a ball vertically upwards with speed of $10\,\mathrm{m/s}$. Range of the ball as observed by an observer at rest on the river bank, is \_\_\_\_ cm. (Take $g = 10\,\mathrm{m/s^2}$)