Question

A ball dropped from a window strikes the ground $2.73s$ later. How high is the window above the ground?

Answer 1

The ball is dropped from a window; it travels downwards under the action of gravitational force, so its acceleration is equal to acceleration due to gravity. Substituting corresponding values in the equation of motion, we can calculate the height of the window from the ground.
Formulas used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
Given, a ball is dropped from a window and hence starts from rest and strikes the ground after a time interval of $2.73s$. The ball comes down under the action of gravitational force and hence is equal to the acceleration due to gravity, i.e. $a=10m{{s}^{-2}}$.
Since the acceleration of the ball is constant, we can use the following equation of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Here, $s$ is the displacement
$u$ is the initial velocity
$t$ is the time taken
$a$ is the acceleration
Substituting given values in the above equation we get,
$\begin{aligned} & s=0+\dfrac{1}{2}\times 10\times {{(2.73)}^{2}} \\\ & \Rightarrow s=\dfrac{75}{2} \\\ & \therefore s=34.5m \\\ \end{aligned}$
The distance travelled by the ball is $34.5m$.
Therefore, the window is $34.5m$ from the ground.

Additional Information:
The equations of motions are used to describe the motion of a body. It gives us the relation between initial velocity, final velocity, displacement travelled, time taken and acceleration of the body and are applied when the acceleration of the body is constant. The different equations of motion are; $v=u+at$, ${{v}^{2}}={{u}^{2}}+2as$, $s=ut+\dfrac{1}{2}a{{t}^{2}}$.

Note:
The motion of the ball is in vertical direction; hence the equation of motion is applied in the vertical direction. According to Newton’s second law of motion, the force is related to mass and acceleration. Therefore, if the acceleration is constant, the force acting on the body is constant.