Answer to Question #205025 in Mechanics | Relativity for Asha

Question

A projectile is projected with a velocity u such that it's range is twice the greatest height attached then it's range is given by

Answer 1

Explanations & Calculations

• Assume it is projected at an angle $\small \theta$ above from the horizontal.
• Then the range could be written as,

$\qquad\qquad \begin{aligned} \small R&=\small \frac{u^2 \sin2 \theta}{g} \end{aligned}$

• And the maximum height as,

$\qquad\qquad \begin{aligned} \small H&=\small \frac{u^2 \sin^2\theta }{2g} \end{aligned}$

• According to the data,

$\qquad\qquad \begin{aligned} \small R&=\small 2H\\ \frac{u^2\sin2 \theta}{g}&=\small \frac{u^2\sin ^2\theta}{g}\\ \small 2\sin\theta\cos\theta&=\small \sin^2\theta\\ \small \sin\theta (2\cos\theta-\sin\theta)&=\small 0\\\\ \small 2\cos\theta-\sin\theta&=\small 0\\ \small \tan\theta&=\small 2 \end{aligned}$

• Now the value of the tangent of the projection angle is known & what is necessary is that the value of $\small \sin 2\theta$.
• This can be written in terms of tangents.
• Then the value of it can be found & the value for the range thereafter.

$\qquad\qquad \begin{aligned} \small \sin 2\theta&=\small \frac{2\tan \theta}{1+\tan^2\theta}\\ &=\small \frac{4}{5}\\\\ \small \therefore R&=\small \bold{\frac{4u^2}{5g}} \end{aligned}$