Question:

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $ 30{}^\circ $ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $ (g=10\,m/{{s}^{2}},\,\sin {{30}^{o}}=1/2,\,\cos {{30}^{o}}=\sqrt{3}/2) $

Updated On: Jul 5, 2022
  • 5.20m
  • 4.33m
  • 2.60m
  • 8.66m
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The Correct Option is D

Solution and Explanation

The ball will be at point $ P $ when it is at a height of $ 10\,\,m $ from the ground. So, we have to find distance $ OP $ , which can be calculated direct considering it as a projectile on a levelled $ (OX) $ . $ OP=R=\frac{{{u}^{2}}\sin 2\theta }{g} $ $ =\frac{{{10}^{2}}\times \sin (2\times {{30}^{o}})}{10} $ $ =\frac{10\sqrt{3}}{2}=5\sqrt{3} $ $ =8.66\,\,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