TS EAMCET · Physics · Motion In Two Dimensions
If a body projected with a velocity of \(19.6 \mathrm{~ms}^{-1}\) reaches a maximum height of 9.8 m, then the range of the projectile is
(Neglect air resistance)
- A 19.6 m
- B 39.2 m
- C 78.4 m
- D 9.8 m
Answer & Solution
Correct Answer
(B) 39.2 m
Step-by-step Solution
Detailed explanation
\(\sin^2 \theta = \frac{2gH}{u^2} = \frac{2 \times 9.8 \times 9.8}{(19.6)^2} = 0.5\) \(\sin \theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ\) \(R = \frac{u^2 \sin(2\theta)}{g} = \frac{(19.6)^2 \sin(2 \times 45^\circ)}{9.8} = \frac{384.16 \times 1}{9.8} = 39.2 \mathrm{~m}\)
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