Question

: A particle is projected vertically upwards from a point $A$ on the ground. It takes time ${t_1}$ to reach a point $B$, but it continues to move up. If it takes further time ${t_2}$ to reach the ground from the point $B$, then the height of the point $B$ from the ground is:

Answer 1

Hint We will use the equations of motion for solving the above problem. Here the general direction of gravity $g$, is assumed to be negative so that the value of acceleration in the equations of motion is $- g$. The equation of motion i.e. $v = u + at$ is to be applied at a time ${t_1} + {t_2}$, i.e. at the end of the motion and the displacement equation $S = ut + \dfrac{1}{2}a{t^2}$, is to be applied at the time ${t_2}$ when the particle is at the point $B$.
Formula used Equations of motion namely $v = u + at$, and $S = ut + \dfrac{1}{2}a{t^2}$.

Complete Step by step solution We will use the general laws of motion to do this task. Since the particle is projected vertically upwards, it has an initial velocity $u$.
It will have the same magnitude of velocity in the opposite direction when it falls after the entire motion is completed. Thus, the final velocity of the particle at the time ${t_1} + {t_2}$ is $- u$.
Using these values in the general equation of motion, we get
$- u = u + ( - g)({t_1} + {t_2})$.
$\Rightarrow 2u = g({t_1} + {t_2})$, or $u = \dfrac{1}{2}g({t_1} + {t_2})$.
Using this value of initial velocity $u$ in the displacement equation of motion, we will get the following equation,
$h = u{t_1} - \dfrac{1}{2}g{t_1}^2$,
where $h$ is the required height of the point $B$.
Using the above value of $u = \dfrac{1}{2}g({t_1} + {t_2})$ in the displacement equation, we will get, $h = (\dfrac{1}{2}g({t_1} + {t_2})){t_1} - \dfrac{1}{2}g{t_1}^2$
$\Rightarrow h = \dfrac{1}{2}g{t_1}^2 + \dfrac{{g{t_1}{t_2}}}{2} - \dfrac{1}{2}g{t_1}^2$
Therefore we get $h = \dfrac{{g{t_1}{t_2}}}{2}$.

This is the required solution and the height of the point $B$.

Note We can also use the three equations to solve for the same problem as given below, $- u = u - g({t_1} + {t_2})$, $v = u - g{t_1}$ and ${v^2} - {u^2} = 2( - g)h$. Substituting $u$ and $v$ from the first two equations into the third equation, we will get the same value of the required height $h$.