$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

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 $ms^{-2}$)

$A$ ball is thrown upwards and it returns to the ground describing a parabolic path. Which of the following remains constant?

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

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

$A$ body is projected with a velocity $(\hat{i} + 2\hat{j}) \text{ ms}^{-1}$,where $\hat{i}$ is along the horizontal and $\hat{j}$ is vertically upward. Then the equation of its trajectory is $(g = 10 \text{ ms}^{-2})$.