Question

If we throw a body upwards with a velocity of $4m/s$, at what height does its kinetic energy reduce to half of the initial value (Taking $g=10m/s$)?

• A. $4m$
• B. $2m$
• C. $1m$
• D. $0.4m$

Answer 1

The correct option is D

$0.4m$

Step 1: Given

Velocity of the body: $v=4m/s$

Acceleration due to gravity: $g=10m/s$

Step 2: Formula Used

Kinetic Energy: $KE=\frac{1}{2}m{v}^2$

Potential Energy: $PE=mgh$

Step 3: Find the required value of height

Calculate the change in kinetic energy by subtracting the initial kinetic energy from the final kinetic energy (half of the initial kinetic energy).

$\begin{array}{rcl}∆KE& =& K{E}_{initial}-K{E}_{final}\\ & =& \frac{1}{2}m{\left(4\right)}^2-\frac{\frac{1}{2}m{\left(4\right)}^2}{2}\\ & =& 8m-4m\\ & =& 4m\end{array}$

According to the law of conservation of energy, change in kinetic energy is equal to change in potential energy. Equate the change in kinetic energy with potential energy (as the initial value of the potential energy is zero) and solve for height.

$$\begin{gathered} PE=∆KE \\ \Rightarrow mgh=4m \\ \Rightarrow m\times 10\times h=4m \\ \Rightarrow h=\frac{4}{10} \\ \Rightarrow h=0.4 \end{gathered}$$

Thus, the height at which kinetic energy will remain half of its initial value is $0.4m$.

Hence, the option D is correct.