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
$\frac{\sqrt{3}}{2}$
$2$
$\frac{1}{2}$
$\frac{2}{\sqrt{3}}$
A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)
A boy throws a ball in air at $60^o$ to the horizontal, along a road with a speed of $10\,ms^{-1}$ $(36\, km/h)$. Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of $(18\, km/h)$. Give explanation to support your diagram.
A player kicks a football with an initial speed of $25\, {ms}^{-1}$ at an angle of $45^{\circ}$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g $=10 \,{ms}^{-2}$ )
The projectile motion of a particle of mass $5\, g$ is shown in the figure.
The initial velocity of the particle is $5 \sqrt{2}\, ms ^{-1}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is $x \times 10^{-2}\, kgms ^{-1} .$ The value of $x ,$ to the nearest integer, is ...... .
A hill is $500\, m$ high. Supplies are to be sent across the hill, using a canon that can hurl packets at a speed of $125 \,m/s$ over the hill. The canon is located at a distance of $800 \,m$ from the foot of hill and can be moved on the ground at a speed of $2\, ms^{-1}$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take, $g = 10\, ms^{-2}$.