Question

A stone lying at the edge of a road falls down into a river flowing below the road. If the stone takes $2$ seconds to reach the river, the find (i) the speed with which the stone strikes the water surface in the river, (ii) the average speed of the stone at the end of the fall and (iii) the height of the road from the river. $(Take,g=9.8m{s}^{-2})$

Answer 1

Step 1:Given data

Time,$t=2s$

Initial velocity, $u=0$

Acceleration due to gravity, $g=9.8m/s^2$

Step 2: Finding the velocity

The speed with which stones strike the water surface in the rivers is $v=u+gt$

Substituting the values into the formula

$v=0+2\times 9.8$

$v=19.6m{s}^{-1}$

Therefore, the speed with which the stone strikes the water in the river is $19.6m{s}^{-1}$.

Step 3: Finding the average speed

Since acceleration of the stone is constant;

Average speed$=\frac{u+v}{2}=\frac{0+19.6}{2}$

Average speed$=9.8m{s}^{-1}$

Step 4: Finding the height of the road from the river

The height of the road from the river is given by

$s=ut+\frac{1}{2}g{t}^2$

$=0\times 2+\frac{1}{2}\times 9.8\times {\left(2\right)}^2$

$=0+9.8\times 2$

$=19.6m$

So, the height of the road from the river is $19.6m$.