$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

$A$ cricket fielder can throw a cricket ball with a speed $v_{0}$. If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal,find:
$(a)$ The effective angle to the horizontal at which the ball is projected in the air as seen by a spectator.
$(b)$ The time of flight.
$(c)$ The horizontal range from the point of projection at which the ball will land.
$(d)$ The angle $\theta$ at which he should throw the ball to maximize the horizontal range found in $(c)$.
$(e)$ How does $\theta$ for maximum range change if $u > v_{0}$,$u = v_{0}$,and $u < v_{0}$?
$(f)$ How does $\theta$ in $(e)$ compare with that for $u = 0$ (i.e.,$45^{\circ}$)?

The angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)

Path of projectile is given by the equation $Y = P x - Q x^2$, match the following accordingly (acceleration due to gravity = $g$)

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

$A$ helicopter flying horizontally with a velocity of $288 \ km/h$ drops a bomb. If the line joining the point of dropping the bomb and the point where the 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 $= 10 \ m/s^2$) (in $m$)

Two towers $A$ and $B$,each of height $20 \ m$,are situated a distance $200 \ m$ apart. $A$ body thrown horizontally from the top of the tower $A$ with a velocity $20 \ ms^{-1}$ towards the tower $B$ hits the ground at point $P$,and another body thrown horizontally from the top of tower $B$ with a velocity $30 \ ms^{-1}$ towards the tower $A$ hits the ground at point $Q$. If a car starting from rest from $P$ reaches $Q$ in $10 \ s$,then the acceleration of the car is (acceleration due to gravity $g = 10 \ ms^{-2}$): (in $ms^{-2}$)