A ball is projected at an angle $45^o$ with horizontal. It passes through a wall of height $h$  at horizontal distance $d_1$ from the point of projection and strikes the ground at a  horizontal distance $(d_1 + d_2)$ from the point of projection, then $h$ is

A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle ' $\theta$ ' to the horizontal, the maximum height attained by it equals $4 \mathrm{R}$. The angle of projection, $\theta$, is then given by

The range of a projectile when launched at angle $\theta$ is same as when launched at angle $2 \theta$. What is the value of $\theta$ ?

A projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$. At the same time another projectile $B$ is thrown with velocity $v_2$ upwards from the point $Q$ vertically below the highest point $A$ would reach. For $B$ to collide with $A$, the ratio $\frac{v_2}{v_1}$ should be

Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.