MCQ
A stone is projected at angle $30^{\circ}$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be :
- A$1: 2$
- B$1: 4$
- C$4: 1$
- ✓$4: 3$
✓
Answer
Correct option: D.
$4: 3$
d
$\frac{ KE _{ POP }}{ KE _{\text {top }}}=\frac{\frac{1}{2} M ( u )^2}{\frac{1}{2} M \left( u \cos 30^{\circ}\right)^2}=\frac{4}{3}$
$\frac{ KE _{ POP }}{ KE _{\text {top }}}=\frac{\frac{1}{2} M ( u )^2}{\frac{1}{2} M \left( u \cos 30^{\circ}\right)^2}=\frac{4}{3}$
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