$A$ body when projected at an angle $\theta$ with the horizontal reaches a maximum height $H$. The time of flight of the body will be ($g=$ acceleration due to gravity).

$A$ body of mass $M$ thrown horizontally with velocity $v$ from the top of a tower of height $H$ touches the ground at a distance of $100 \ m$ from the foot of the tower. $A$ body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of a tower of height $4H$ will touch the ground at a distance of ........ $m$.

$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.

$A$ particle is thrown with a velocity of $u \, m/s$ at an angle of $30^{\circ}$ with the horizontal. It passes through points $A$ and $B$ at the same vertical height at times $t_1 = 1 \, s$ and $t_2 = 3 \, s$ respectively,as shown in the figure. The value of $u$ is ....... $m/s$ (take $g = 10 \, m/s^2$).

$A$ 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.,$45^{\circ}$)?

$A$ body is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $60 \sqrt{2} \ m/s$. Then the angle made by its velocity with the horizontal after $6 \ s$ is (in $^{\circ}$)