Question

A arrow is straight up at an initial velocity of $250m{{s}^{-1}}$. How long will it take to hit the ground?

Answer 1

Hint: The arrow is shot straight vertically upwards. It reaches a height and comes back to the ground under the action of the gravitational force. Using the equation of motion which gives the relation between displacement, initial velocity, time taken and acceleration and substituting the corresponding values, we can calculate the time taken.
Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
The arrow is shot vertically upwards and it traces its path under the action of force of gravity. Therefore, the acceleration of the arrow is acceleration due to gravity, $10m{{s}^{-2}}$.
As the force acting on the arrow is constant, we can use equations of motion to analyse its motion
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Here, $s$ is the displacement
$u$ is the initial velocity
$a$ is the acceleration
$t$ is the time taken
It started from the ground and if it ends back to the ground, the displacement will be 0. Given, initial velocity is $250m{{s}^{-1}}$. We substitute given values in the above equation, to get,
$\begin{align}
  & 0=250\times t+(-10){{t}^{2}} \\
 & \Rightarrow 250t-10{{t}^{2}}=0 \\
 & \Rightarrow 10t(25-t)=0 \\
 & \therefore t=0,\,t=25s \\
\end{align}$
The arrow will have zero displacement at two instances, initially at the beginning and finally at $t=25s$.
Therefore, the arrow will take $25s$ to hit the ground.

Note:
The equations of motion describe the motion of a body in one dimension. According to the second law of motion, the force is directly proportional to the acceleration. Since the force is constant, acceleration is also constant. The kinetic energy is maximum closest to the ground while the potential energy is maximum at the highest point. A quadratic equation has two roots and hence the arrow has zero displacement at two time intervals.