Question:

If $R$ and $H$ represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Updated On: Jul 6, 2022
  • $tan^{-1}\left(\frac{H}{R}\right)$
  • $tan^{-1}\left(\frac{2H}{R}\right)$
  • $tan^{-1}\left(\frac{4H}{R}\right)$
  • $tan^{-1}\left(\frac{4R}{H}\right)$
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The Correct Option is C

Solution and Explanation

Maximum height, $H=\frac{u^{2}\,sin^{2}\,\theta}{2g}\,...\left(i\right)$ Horizontal range, $R=\frac{u^{2}\,sin^{2}\,2\theta }{g}$ $=\frac{2u^{2}\,sin\,cos\,\theta}{g}\,...\left(ii\right)$ Divide $\left(i\right)$ by $\left(ii\right)$, we get $\frac{H}{R}=\frac{tan\,\theta}{4}$ or $tan\,\theta =\frac{4H}{R}$ or $\theta=tan^{-1}\left(\frac{4H}{R}\right)$
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration