Search any question & find its solution
Question:
Answered & Verified by Expert
A body of mass $2 \mathrm{~kg}$ is projected vertically upwards with a velocity of $2 \mathrm{~m~} \mathrm{sec}^{-1}$. The K.E. of the body just before striking the ground is
Options:
Solution:
2930 Upvotes
Verified Answer
The correct answer is:
$4 \mathrm{~J}$
If particle is projected vertically upward with velocity of $2 \mathrm{~m} / \mathrm{s}$ then it returns with the same velocity.
So its kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times(2)^2=4 \mathrm{~J}$
So its kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times(2)^2=4 \mathrm{~J}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.