$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
Choose the $CORRECT$ answer.

$A$ projectile is thrown such that it achieves a maximum range $R$. What are the coordinates of the point where its velocity is minimum?

$A$ particle is projected in air at some angle to the horizontal and moves along a parabola as shown in the figure,where $x$ and $y$ indicate horizontal and vertical directions,respectively. The diagram shows the direction of velocity and acceleration at points $A$,$B$,and $C$.

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is: (projection speed = $u$,angle of projection from horizontal = $\theta$)

$A$ boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle of $30^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\, m$ from the ground? $\left[ g = 10\, m/s^2, \sin 30^o = \frac{1}{2}, \cos 30^o = \frac{\sqrt{3}}{2} \right]$