A projectile is thrown with an initial velocity of $(a \hat{ i }+b \hat{ j }) ms ^{-1}$. If the range of the projectile is twice the maximum height reached by it, then

At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown The distance (in metres) of the particle from origin at $t = 2$ sec.

A gun can fire shells with maximum speed $v_0$ and the maximum horizontal range that can be achieved is $R_{max} = \frac {v_0^2}{g}$.  If a target farther away by distance $\Delta x$ (beyond $R$) has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least $h = \Delta x\,\left[ {1 + \frac{{\Delta x}}{R}} \right]$.

Show that for a projectile the angle between the velocity and the $x$ -axis as a function of time is given by

$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by

$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$

Where the symbols have their usual meaning.

A person can throw a ball upto a maximum range of $100 \,m$. How high above the ground he can throw the same ball?

A particle is thrown with a velocity of $u \,m / s$. It passes $A$ and $B$ as shown in figure at time $t_1=1 \,s$ and $t_2=3 \,s$. The value of $u$ is ....... $m / s$ $\left(g=10 \,m / s ^2\right)$