$A$ stone is thrown vertically at a speed of $30 \,ms^{-1}$ making an angle of $45^{\circ}$ with the horizontal. What is the maximum height reached by the stone (in $\,m$)? Take $g=10 \,ms^{-2}$.

$A$ 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 $g=10 \,m/s^2$):

$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

$A$ fighter jet is flying horizontally at a certain altitude with a speed of $200 \; m/s$. When it passes directly overhead an anti-aircraft gun,a bullet is fired from the gun at an angle $\theta$ with the horizontal to hit the jet. If the bullet speed is $400 \; m/s$,the value of $\theta$ will be $\dots \; ^\circ$.

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)