Question

A stone projected at an angle $\theta$ with horizontal from the roof of a tall building falls on the ground after three seconds. Two seconds after the projection it was again at the level of projection. Then the height of the building is:

Answer 1

The height of the building is determined by using the height of the projectile motion by using the time given in the question. First by using the time equation and then by using the height equation, then the height of the building is determined.

Useful formula
The time taken for the projectile motion is given by,
$T = \dfrac{{2u\sin \theta }}{g}$
Where, $T$ is the time taken, $u$ is the velocity of the object, $\theta$ is the angle of the object is thrown and $g$ is the acceleration due to gravity.
The height of the object in projectile motion is given by,
$H = ut + \dfrac{1}{2}a{t^2}$
Where, $H$ is the height of the object, $u$ is the velocity of the object $t$ is the time taken by the object and $a$ is the acceleration of the object.

Complete step by step solution
Given that,
The angle of the projection of the stone is, $\theta$
The time taken by the stone to fall on the ground is, $t = 3\,\sec$
The time taken by the stone to the level of projection is, $T = 2\,\sec$
Now,
The time taken for the projectile motion is given by,
$T = \dfrac{{2u\sin \theta }}{g}\,......................\left( 1 \right)$
Here, assume that the acceleration due to gravity is equal to $10\,m{s^{ - 2}}$ and substituting the value of the time, then
$2 = \dfrac{{2u\sin \theta }}{{10}}$
By cancelling the same terms on both sides, then the above equation is written as,
$1 = \dfrac{{u\sin \theta }}{{10}}$
By rearranging the terms in the above equation, then
$u\sin \theta = 10$
Now,
The height of the object in projectile motion is given by,
$H = ut + \dfrac{1}{2}a{t^2}\,......................\left( 2 \right)$
By substituting the velocity of the stone and the time taken by the stone and the acceleration due to gravity in the above equation, then the above equation is written as,
$H = \left( { - u\sin \theta \times 3} \right) + \dfrac{1}{2}\left( {10 \times {3^2}} \right)$
Here the negative sign indicates that the height of the building is below the position of the stone thrown.
By substituting the value of $u\sin \theta$ in the above equation and squaring the term in the above equation, then
$H = - \left( {10 \times 3} \right) + \dfrac{1}{2}\left( {10 \times 9} \right)$
By multiplying the terms in the above equation, then
$H = - 30 + \dfrac{1}{2}\left( {90} \right)$
On dividing the terms in the above equation, then
$H = - 30 + 45$
By adding the terms in the above equation, then the above equation is written as,
$H = 15\,m$

Thus, the height of the building is $15\,m$.

Note: The negative sign indicates that the stone will travel below the position of the stone is thrown. First, by using the time equation the velocity is determined and using that velocity in the height formula, then the height of the building is determined.