A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ............... m. $\left(\right.$ Use $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$.
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & v^2=u^2+2 a s \\ & 100=0+2(10) s \\ & S=5 m \\ & \text { Height from ground }=10-5=5 m\end{aligned}$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
