$A$ train is moving along a straight line with a constant acceleration '$a$'. $A$ boy standing in the train throws a ball forward with a speed of $10 \ m/s$,at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 \ m$ inside the train to catch the ball back at the initial height. The acceleration of the train,in $m/s^2$,is

Read each statement below carefully and state,with reasons,if it is true or false:
$(a)$ The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
$(b)$ The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
$(c)$ The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

$A$ particle moves with constant angular velocity in a circle. During the motion,its:

$A$ ball of mass $m$ is thrown vertically upward. Another ball of mass $2m$ is thrown at an angle $\theta$ with the vertical. Both the balls stay in the air for the same period of time. The ratio of the maximum heights attained by the two balls respectively is $\frac{1}{x}$. The value of $x$ is $.....$

$A$ particle leaves the origin with an initial velocity $\vec{v} = (3 \hat{i}) \text{ m s}^{-1}$ and a constant acceleration $\vec{a} = (-1 \hat{i} - 0.5 \hat{j}) \text{ m s}^{-2}$. The position vector of the particle, when it reaches its maximum $x$-coordinate, is:

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)