$A$ 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

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

$A$ projectile is launched from the ground, such that it hits a target on the ground which is $90 \,m$ away. The minimum velocity of the projectile to hit the target is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-1}$)

$A$ 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 $.........\,m$.

$A$ stone projected from the ground with a velocity $50 \ m/s$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of $3 \ s$. Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $g = 10 \ m/s^2$) (in $m$)

Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.