$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$,where $h$ is the height,$x$ is the horizontal distance,and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

$A$ small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $v_0$. The velocity is averaged for the first $\sqrt{2} \,s$ and the magnitude of the average velocity comes out to be the same as that of the initial velocity,i.e.,$|v_0|$. The magnitude $|v_0|$ will be (take $g=10 \,m/s^2$):

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 = $\theta$)