(1)

Given, $u=4 {\text{ms}}^{-1}, v=0, g=10 {\text{ms}}^{-2}$

Time taken by ball to reach its maximum height,

     $v=u+at$

$\Rightarrow 0=4+(-10)t \text{or} 4=10t$

      $t=0.4 \text{sec}.$

The height of bridge above water surface is

$S=ut+\frac{1}{2}a{t}^2\Rightarrow S=\left(0\right)t+\frac{1}{2}\left(10\right){(3.6)}^2$

$S=64 \text{m}$

(3)

Distance travelled by a body during free fall is given by

$S=ut+\frac{1}{2}a{t}^2$

Here, $u=0$ and $a=g$

$\therefore$     $S\propto t^2$

Here for 1st second,

          $S_1=K{t}^2=K$

For 2nd second,

          $S_2=K{\left(2\right)}^2=4K$

For 3rd second,

          $S_3=K{\left(3\right)}^2=9K$

For 4th second,

          $S_4=K{\left(4\right)}^2=16K$

Distance covered in 2nd seconds, $4K-K=3K$

Distance covered in 3rd second, $9K-4K=5K$

Distance covered in 4th second, $16K-9K=7K$

$\therefore$    Ratio of distances travelled by the freely falling body will be $1:3:5:7$

(4)

Here, $u=20 \text{m/s}, v=80 \text{m/s}, g=10 {\text{m/s}}^2, h=?$

$v^2=u^2+2gh\Rightarrow {80}^2={20}^2+2\times 10\times h.$ Hence, $h=300 \text{m}$