Referring to above question, the angle with the horizontal at which the projectile was projected is

1. $ta{n}^{-1}\left(\frac{3}{4}\right)$
2. $ta{n}^{-1}\left(\frac{4}{3}\right)$
3. $Notobtainablefromthegivendata$
4. $si{n}^{-1}\left(\frac{3}{4}\right)$

A particle is projected from ground level with an initial velocity of $35m/s$ at an angle of ${\tan }^{-1}3/4$ to the horizontal. Find the time for which the particle is more than $2m$ above the ground. $[g=10m/s^2]$

A ball is projected at an angle of ${30}^0$ above with the horizontal from the top of a tower and strikes the ground in $5$ sec at an angle of ${45}^0$ with the horizontal. Find the height of the tower and the speed with which it was projected.

For a projectile projected from ground at an angle $\theta$ with horizontal, $g{T}^2=\frac{2R}{\sqrt{3}}$, where $T$ is time of fight, $R$ is horizontal range of projectile, $g$ is acceleration due to gravity. The angle of projection $\left(\theta \right)$ is

Referring to above question, the angle with the horizontal at which the projectile was projected is

A particle is projected from ground level with an initial velocity of $35m/s$ at an angle of ${\tan }^{-1}3/4$ to the horizontal. Find the time for which the particle is more than $2m$ above the ground. $[g=10m/s^2]$

A ball is projected at an angle of ${30}^0$ above with the horizontal from the top of a tower and strikes the ground in $5$ sec at an angle of ${45}^0$ with the horizontal. Find the height of the tower and the speed with which it was projected.

For a projectile projected from ground at an angle $\theta$ with horizontal, $g{T}^2=\frac{2R}{\sqrt{3}}$, where $T$ is time of fight, $R$ is horizontal range of projectile, $g$ is acceleration due to gravity. The angle of projection $\left(\theta \right)$ is