Question

For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms^–2)

• A. 5 : 4
• B. 5 : 2
• C. 5 : 1
• D. 10 : 1

Answer 1

Answer: (A)

5 : 4

Answer 2

$H = \frac{u^{2}\sin^{2}\theta}{2g}$and $T = \frac{2u\sin\theta}{g}$

∴ $\frac { H } { T ^ { 2 } } = \frac { u ^ { 2 } \sin ^ { 2 } \theta / 2 g } { 4 u ^ { 2 } \sin ^ { 2 } \theta / g ^ { 2 } }$

$= \frac{g}{8} = \frac{10}{8} = \frac{5}{4}$