$A$ projectile is given an initial velocity of $\hat{i}+2 \hat{j} \,ms^{-1}$. The Cartesian equation of its path is ($x$ and $y$ are in metres and $g=10 \,ms^{-2}$)

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

$A$ body is projected horizontally from the top of a tower with an initial velocity of $18 \, m s^{-1}$. It hits the ground at an angle of $45^{\circ}$. What is the vertical component of velocity when it strikes the ground? $(ms^{-1})$

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then