Solveeit Logo
Login

Question

Question: A particle is projected with a velocity v such that its range on the horizontal plane is twice the g...

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

A

4v25g\frac{4v^{2}}{5g}

B

4g5v2\frac{4g}{5v^{2}}

C

v2g\frac{v^{2}}{g}

D

FA, 6muFBF_{A},\mspace{6mu} F_{B}

Answer

4v25g\frac{4v^{2}}{5g}

Explanation

Solution

We know R=4Hcot⁡θR = 4H\cot\theta

2H=4Hcot⁡θ2H = 4H\cot\theta ⇒ cot⁡θ=12\cot\theta = \frac{1}{2}; sin⁡θ=25\sin\theta = \frac{2}{\sqrt{5}}; cos⁡θ=15\cos\theta = \frac{1}{\sqrt{5}} [As R=2H given]\left\lbrack \text{As }R = 2H\text{ given} \right\rbrack

Range=u2.2.sin⁡θ.cos⁡θg\text{Range} = \frac{u^{2}.2.\sin\theta.\cos\theta}{g} =2u225.15g= \frac{2u^{2}\frac{2}{\sqrt{5}}.\frac{1}{\sqrt{5}}}{g} =4u25g= \frac{4u^{2}}{5g}