MCQ
For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10 ms^{-2}$)
- ✓$5:4$
- B$5:2$
- C$5:1$
- D$10:1$
✓
Answer
Correct option: A.
$5:4$
a
(a) $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$
(a) $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$
So $\frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}\theta /2g}}{{4{u^2}{{\sin }^2}\theta /{g^2}}} = \frac{g}{8} = \frac{5}{4}$
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