Question:

An arrow is shot into air. Its range is 200 m and its time of flight is 5 s. If g = 10 $ ms^{- 2} $, then the horizontal component of velocity of the arrow is

Updated On: Jul 5, 2022
  • $ 12.5 \, ms^{ - 1} $
  • $ 25 \, ms^{ - 1} $
  • $ 31.25 \, ms^{ - 1} $
  • $40 \, ms^{ - 1} $
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The Correct Option is D

Solution and Explanation

T = $ \frac{ 2 u \sin \theta }{ g } $ $ \hspace20mm $ ..(i) Also horizontal range R = $ u \cos \, \theta \times t $ R = $ \frac{ u \, \cos \, \theta \times 2 u \, \sin \, \theta }{ g } $ $ \hspace20mm $ ..(ii) Given T = S and R = 200 $ \therefore $ From Eqs. (i) and (ii), we get 5 = $ \frac{ 2 u \sin \, \theta }{ g } $ $ \hspace20mm $ ..(iii) 200 = $ \frac{ 2 u^2 \, \sin \, \theta \, \cos \, \theta }{ g } $ $ \hspace20mm $ ..(iv) Dividing E (iv) by E (iii), we get $ \frac{ 200 }{ 5} = u \cos \theta = 40 \, ms^{ - 1} $
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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