$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

The angle of projection for a projectile to have the same horizontal range and maximum height is

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)

$A$ body of mass $m$ thrown up vertically with velocity $v_1$ reaches a maximum height $h_1$ in $t_1$ seconds. Another body of mass $2m$ is projected with a velocity $v_2$ at an angle $\theta$. The second body reaches a maximum height $h_2$ in time $t_2$ seconds. If $t_1 = 2t_2$, then the ratio $\left(\frac{h_1}{h_2}\right)$ is

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

What is the range of a projectile? Also,provide the velocity of the projectile at its maximum height.