Two stones are projected from the top of a cliff $h$ metres high, with the same speed $u$, so as to hit the ground at the same spot. If one of the stones is projected at an angle $\theta$ to the horizontal then the $\theta$ equals
(1) $u\sqrt{\frac{2}{gh}}$
(2) $\sqrt{\frac{2u}{gh}}$
(3) $2g\sqrt{\frac{u}{h}}$
(4) $2h\sqrt{\frac{u}{g}}$

A ball is thrown from a point with a speed $v _ { 0 }$ at an angle of projection $\theta$. From the same point and at the same instant person starts running with a constant speed $v _ { 0 } / 2$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?
(1) yes, $60 ^ { \circ }$
(2) yes, $30 ^ { \circ }$
(3) no
(4) yes, $45 ^ { \circ }$

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