$A$ car moving with a certain velocity jumps from an inclined plane placed at one bank of a river and reaches the other bank by attaining a maximum height of $80 \,m$. If the same car, moving with the same velocity, jumps from another inclined plane having a different angle of inclination and reaches the same point on the other bank by attaining a maximum height of $45 \,m$, then the width of the river is: (in $\,m$)

Fill in the blanks:
$(a)$ For vectors $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} + \hat{j} - 2\hat{k}$,the $y$-component of the vector $(\vec{A} - \vec{B})$ is .....
$(b)$ The formula for the total time of flight of a projectile projected vertically upwards with an initial velocity $u$ is ..........
$(c)$ The formula for the centripetal acceleration of a particle moving in a circular path of radius $R$ with a constant angular velocity $\omega$ is ......
$(d)$ The component of any vector is always .........

$A$ cyclist starts from the centre $O$ of a circular park of radius $1 \text{ km}$,reaches the edge $P$ of the park. Then,the cyclist cycles along the circumference to point $Q$ and returns to the centre along $QO$ as shown in the figure. If the round trip takes $10 \text{ minutes}$,the net displacement and average speed of the cyclist (in metre and kilometre per hour respectively) are:

An athlete completes one round of a circular track of radius $R$ in $40 \, s$. What will be his displacement at the end of $2 \, min \, 20 \, s$?

$A$ particle is projected from the ground at an angle $\theta$ with the horizontal with speed $u$. The ratio of the radius of curvature of its trajectory at the point of projection to the radius of curvature at the maximum height is ........

$A$ 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 . . . . . . .