Question:

An artillery piece which consistently shoots its shells with the same muzzle speed has a maximum range $R$ . To hit a target which is $\frac{R}{2}$ from the gun and on the same level, the angle of elevation of the gun should be

Updated On: Jul 2, 2022
  • $15^\circ $
  • $45^\circ $
  • $30^\circ $
  • $60^\circ $
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The Correct Option is A

Solution and Explanation

$\text{R}_{\text{max}} = \text{R} = \frac{\text{u}^{2}}{\text{g}}$ (at an angle of $45^\circ $ $\text{u}^{2} = \text{Rg}$ Using $\text{Range} = \frac{\text{u}^{2} \text{sin} 2 \theta }{\text{g}}$ then $\frac{\text{R}}{2} = \left(\text{Rg}\right) \frac{\text{sin} 2 \theta }{\text{g}}$ $\text{sin} 2 \theta = \frac{1}{2}$ $2 \theta = 3 0^{\text{o}} \, \, \, ? \, \, \, \theta = 1 5^{\text{o}}$
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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