Question:

From the ground, a projectile is fired at an angle of $60$ degrees to the horizontal with a speed of $20\, m/s^{-1}$. Take acceleration due to gravity as $10 \, m/s^{-2}$. The horizontal range of the projectile is

Updated On: Mar 4, 2024
  • $10 \sqrt{3}\, m $
  • 20 m
  • $20 \sqrt{3}\, m $
  • $40 \sqrt{3}\, m $
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The Correct Option is C

Solution and Explanation

The horizontal range
$R=\frac{u^{2} \sin 2 \theta}{g}$
Given, $u=20 \,m / s$
$\theta=60^{\circ}$
$g=10 \,m / s ^{2}$
$R=\frac{(20)^{2} \sin \left(2 \times 60^{\circ}\right)}{10}$
$=\frac{20 \times 20}{10} \times \frac{\sqrt{3}}{2}=20 \sqrt{3} \,m$
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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