$A$ ball is projected from the ground with a speed of $20\,m/s$ at an angle of $45^{\circ}$ with the horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$.

$A$ ball is projected from a building of height $10 \ m$ with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the horizontal. What is the horizontal distance covered by the ball when it reaches the same height of $10 \ m$ again (in $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

$A$ particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in a horizontal direction with a speed of $5 \; m/s$ from the same height. The correct statement is:

$A$ player can throw a ball to a maximum horizontal distance of $80 \,m$. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is (in $\,m$)

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

At the top of the trajectory of a projectile,the acceleration is