Kinematics Question 611
Question: The equation of trajectory of projectile is given by $ y=\frac{x}{\sqrt{3}}-\frac{\text{g}}{{x^{2}}}{20} $ , where x and y are in meter. The maximum range of the projectile is
Options:
A) $ \frac{8}{3}\text{ m} $
B)$ \frac{4}{3}\text{ m} $
C) $ \frac{3}{4}\text{ m} $
D)$ \frac{3}{8}\text{ m} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Comparing the given equation with the equation of trajectory of a projectile, $ y=x\tan \theta -\frac{\text{g}}{{x^{2}}}{2{{u^{2}}}{{\cos }^{2}}\theta }, $ we get, $ \text{tan}\theta \text{=}\frac{1}{\sqrt{3}}\Rightarrow \theta =30{}^\circ $ and $ 2{{u^{2}}}{{\cos }^{2}}\theta =20\Rightarrow {{u^{2}}}=\frac{20}{2{{\cos }^{2}}\theta }=\frac{40}{3} $ Now, $ {{\text{R}} _{\max }}=\frac{{{u _{2}}}}{\text{g}}=\frac{40}{3\times 10}=\frac{4}{3}\text{m} $
BETA
