Question

A stone is thrown in a vertically upward direction with a velocity of $5m{{s}^{-1}}$. If the acceleration of the stone during its motion is $10m{{s}^{-2}}$ in the downward direction, what will be the height attained by the stone and how much time will it reach to take there?
(A). Height= $1.25m$ and time= $0.5s$
(B). Height= $2.5m$ and time= $0.5s$
(C). Height= $2.5m$ and time= $0.05s$
(D). Height= $1.25m$ and time= $0.05s$

Answer 1

Hint: A stone is thrown vertically upwards due to which force of gravity acts on it and an acceleration acts on it in the downward direction. The equation of motion in one dimension gives us the relation between various parameters of a motion like initial velocity, final velocity, acceleration etc. Applying equations of motion for the stone, we can calculate height and time taken.
Formulas used:
${{v}^{2}}={{u}^{2}}-2as$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
The stone is travelling in the vertically upward direction. It will attain a maximum height after which it reverses the direction in the downward direction. We can apply the following equation of motion for its motion in the upward direction
${{v}^{2}}={{u}^{2}}-2as$
$s$ is the displacement travelled
$v$ is the final velocity
$u$ is the initial velocity
$a$ is the acceleration
At the maximum height, the velocity becomes zero, hence, $v=0$. Given, $u=5m{{s}^{-1}}$, $a=-10m{{s}^{-2}}$
In the above equation, we substitute given values to get,
$\begin{align}
  & {{v}^{2}}={{u}^{2}}-2as \\
 & \Rightarrow 0={{(5)}^{2}}-2\times 10\times s \\
 & \Rightarrow 25=20s \\
 & \therefore s=1.25m \\
\end{align}$
The maximum height that the stone attains is $1.25m$.
The time taken by the stone to reach maximum height can be calculated using the equation-
$\begin{align}
  & s=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow 1.25m=5\times t-\dfrac{1}{2}\times 10{{t}^{2}} \\
 & \Rightarrow 1.25=5t-5{{t}^{2}} \\
 & \Rightarrow 0.25=t-{{t}^{2}} \\
 & \Rightarrow 4{{t}^{2}}-4t+1=0 \\
\end{align}$
Applying middle terms split method to solve the quadratic equation we get,
$\begin{align}
  & \Rightarrow 4{{t}^{2}}-2t-2t-1=0 \\
 & \Rightarrow 2t(2t-1)-1(2t-1)=0 \\
 & \Rightarrow (2t-1)(2t-1)=0 \\
 & \Rightarrow t=\dfrac{1}{2} \\
 & \therefore t=0.5s \\
\end{align}$
Therefore, the time taken to reach the height is $0.5s$.
Therefore, the maximum height attained is $1.25m$ and the time taken is $0.5s$.

Hence, the correct option is (A).

Note:
Acceleration is negative because it is acting in the downward direction. The acceleration acting on the stone i.e. $10m{{s}^{-2}}$ is equal to acceleration due to gravity. We can also solve the quadratic equation by using determinant formula or by the squaring method. At maximum height, the potential energy of a body is maximum.