Kinematics Question 619
Question: A particle is projected with a velocity v such that it’s range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g it’s acceleration due to gravity)
Options:
A) $ \frac{4v^{2}}{5g} $
B)$ \frac{4g}{5v^{2}} $
C) $ \frac{v^{2}}{g} $
D)$ \frac{4v^{2}}{\sqrt{5}g} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We know, $ R=4H\cot \theta \Rightarrow \cot \theta =\frac{1}{2} $
$ \text{From triangle we can say that } $
$ \sin \theta =\frac{2}{\sqrt{5}},\text{ cos}\theta \text{=}\frac{1}{\sqrt{5}} $
$ \therefore $ Range of projectile $ R=\frac{2v^{2}\sin \theta \cos \theta }{g} $
$ =\frac{2v^{2}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4v^{2}}{5g} $
BETA
