The speed at the maximum height of a projectile is half of its initial speed $u$. Its range on the horizontal plane is:

The velocity of projection of a projectile is $\left(6\hat{\mathrm{i}}+8\mathrm{j}\right) \mathrm{m} {\mathrm{s}}^{-1}$. The horizontal range of the projectile is $\left(g=10 \mathrm{m} {\sec }^{-2}\right)$

 An object is thrown horizontally from a point $A$ from a tower and hits the ground $3 \mathrm{s}$ later at $B$. The line from $A$ to $B$ makes an angle of $30°$ with the horizontal. The initial velocity of the object is: (take $\left.g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

A body is projected horizontally from the top of a tower with an initial velocity $18 \mathrm{m} {\mathrm{s}}^{-1}$. It hits the ground at angle $45°$. What is the vertical component of velocity when it strikes the ground?

A bomber plane moving at a horizontal speed of $20 \mathrm{m} {\mathrm{s}}^{-1}$ releases a bomb at a height of $80 \mathrm{m}$ above ground as shown. At the same instant, a hunter of negligible height starts running from a point below it, to catch the bomb with speed $10 \mathrm{m} {\mathrm{s}}^{-1}$. After two seconds he realised that he cannot make it, he stops running and immediately holds his gun and fires in such a direction, so that just before the bomb hits the ground, a bullet will hit it. What should be the firing speed of bullet. (Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$ )

A ball is projected from a certain point on the surface of a planet at a certain angle with a horizontal surface. The horizontal and vertical displacement $x$ and $y$ vary with time $t$ in second as:

$x=10\sqrt{3}t$  and $y=10t-t^2$

The maximum height attained by the ball is: