$A$ projectile is thrown with a velocity of $10 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$ with the horizontal. The time interval between the moments when the speed is $\sqrt{125} \ m/s$ is $(g = 10 \ m/s^2)$. (in $s$)

$A$ particle is projected at an angle of $45^{\circ}$ with the horizontal with an initial kinetic energy $K$. What will be its kinetic energy at the highest point?

One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to the horizontal. After two more seconds, it is travelling horizontally. Then the magnitude of the initial velocity of the projectile is $(g = 10 \,ms^{-2})$

Path of projectile is given by the equation $Y = P x - Q x^2$, match the following accordingly (acceleration due to gravity = $g$)

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

$A$ projectile is thrown with a velocity of $50 \, m/s$ at an angle of $53^o$ with the horizontal. Determine the instants at which the projectile is at the same height.

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 = $\theta$)