Question

A person watching through the window of an apartment sees a ball that rises vertically up and then vertically down for a total time of $0 \cdot 5{\text{s}}$ . If the height of the window is $2{\text{m}}$, find the maximum height above the window reached by the ball. Take $g = 10{\text{m}}{{\text{s}}^{ - 2}}$.

Answer 1

Hint:Here it is mentioned that the ball rises and falls vertically. When the ball falls, it can be considered as a free-fall motion of a body. Newton’s first equation of motion for the ball where the maximum height of the ball corresponding to the distance covered by the ball will provide the height above the window.

Formulas used:
-Newton’s first equation of motion gives the distance covered by a body as$s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial velocity of the ball, $t$ is the time taken to cover the distance and $a$ is the acceleration of the body.
-The height from which an object falls under free fall is given by, $h = \dfrac{1}{2}g{t^2}$ where $g$ is the acceleration due to gravity and $t$ is the time taken to reach the ground.

Complete step by step answer.
Step 1: List the given parameters.
The height of the window from which the man sees the motion of the ball is $2{\text{m}}$
Let $h$ be the maximum height above the window and the time taken to cover this height be $t$ .
The total time taken by the ball is given to be $0 \cdot 5{\text{s}}$ .
The time taken to cross the window will be $\left( {t + 0 \cdot 5} \right){\text{s}} = \left( {t + \dfrac{1}{4}} \right){\text{s}}$ .
The distance covered to cross the window will be $\left( {h + 2} \right){\text{m}}$ .
Step 2: Express the equation of motion for the ball to cross the window.
The distance covered by a body is given by, $s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial velocity of the ball, $t$ is the time taken to cover the distance and $a$ is the acceleration of the body.
For the ball, $u = 0$ , $s = \left( {h + 2} \right){\text{m}}$ , $t = \left( {t + \dfrac{1}{4}} \right){\text{s}}$ and $a = g$ .
Then the equation of motion of the ball will be $h + 2 = + \dfrac{1}{2}g{\left( {t + \dfrac{1}{4}} \right)^2}$
Simplifying we get, $h + 2 = + \dfrac{1}{2}g\left( {{t^2} + \dfrac{t}{2} + \dfrac{1}{{16}}} \right)$
Substituting for $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ in the above equation and simplifying we get, $h + 2 = 5{t^2} + \dfrac{{5t}}{2} + \dfrac{5}{{16}}$
$ \Rightarrow h = 5{t^2} + \dfrac{{5t}}{2} - \dfrac{{27}}{{16}}$ -------- (1)
Also as the ball is in free fall, the equation of motion of the ball for the height $h$ can be expressed as $h = \dfrac{1}{2}g{t^2}$ and on substituting for $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ we get,
$h = \dfrac{1}{2} \times 10{t^2} = 5{t^2}$ ---------- (2)
Step 3: Compare equations (1) and (2) to find the height $h$ .
On equating the R.H.S of equations (1) and (2) we get, $5{t^2} + \dfrac{{5t}}{2} - \dfrac{{27}}{{16}} = 5{t^2}$
$ \Rightarrow \dfrac{{5t}}{2} = \dfrac{{27}}{{16}}$
$ \Rightarrow t = \dfrac{{27 \times 2}}{{16 \times 5}} = 0 \cdot 675{\text{s}}$
Thus the time taken to reach the maximum height from the window is $t = 0 \cdot 675{\text{s}}$ .
Now substituting $t = 0 \cdot 675{\text{s}}$ in equation (2) we get, $h = 5 \times {\left( {0 \cdot 675} \right)^2} = 2 \cdot 278{\text{m}}$.
So the maximum height above the window is $h = 2 \cdot 278{\text{m}}$ .

Note:For a body in free fall, its initial velocity will be zero. Also while writing the equation of motion for the ball to cross the window we considered the acceleration to be equal to the acceleration due to gravity. Usually, the acceleration due to gravity is directed downwards and so it will be negative but here we are considering the rising of the ball and so it will be positive.