$A$ car starts from rest and accelerates at $5 \, m/s^{2}$. At $t=4 \, s$,a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t=6 \, s$? (Take $g = 10 \, m/s^{2}$)

$A$ particle of mass $m$ is projected at a velocity $u$, making an angle $\theta$ with the horizontal ($x$-axis). If the angle of projection $\theta$ is varied keeping all other parameters same, then the magnitude of angular momentum $(L)$ at its maximum height about the point of projection varies with $\theta$ as,

Assertion: In a circular motion,work done by centripetal force is not always zero.
Reason: If the speed of the particle increases or decreases in circular motion,then the net force acting on the particle does not remain towards the centre.

$A$ particle starts from the origin at $t=0$ with a velocity $5 \hat{i} \text{ m/s}$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3 \hat{i} + 2 \hat{j}) \text{ m/s}^2$. If the $x$-coordinate of the particle at that instant is $84 \text{ m}$,then the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. The value of $\alpha$ is . . . . . . .

In dealing with the motion of a projectile in the air,we ignore the effect of air resistance on the motion. This gives a trajectory that is a parabola,as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

$A$ projectile is fired from the surface of the earth with a velocity of $5 \, m s^{-1}$ and angle $\theta$ with the horizontal. Another projectile is fired from another planet with a velocity of $3 \, m s^{-1}$ at the same angle and follows a trajectory which is identical to the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $m s^{-2}$) (Given $g = 9.8 \, m s^{-2}$):