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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2019 · Physics · Motion in Two Dimensions

A projectile is given an initial velocity of ( ^ i + 3 ^ j )ms^ - 1...

A projectile is given an initial velocity of \left(\overset{^}{i} + \sqrt{3}\overset{^}{j}\right)ms^{- 1},~where \overset{^}{i} is along the ground and \overset{^}{j} is along the vertical. Then, the equation of the path of projectile is [[Take, \leftg=10ms−2]\leftg = 10ms^{- 2}]

Options

  1. Ay=3x−5x2y = \sqrt{3}x - 5x^{2}
  2. By=3x+5x2y = \sqrt{3}x + 5x^{2}
  3. Cx=3y+5x2x = \sqrt{3}y + 5x^{2}
  4. Dx2=y2+3x^{2} = y^{2} + \sqrt{3}

Correct answer

A. y=3x−5x2y = \sqrt{3}x - 5x^{2}

Step-by-step solution

Velocity of projectile at time tt is given by

v = u - gt\overset{^}{j}

v = \overset{^}{i} + \sqrt{3}\overset{^}{j} - 10t\overset{^}{j}

∴r=∫vdt∴ r = ∫ vdt

= ∫ \left(\overset{^}{i} + \sqrt{3}\overset{^}{j} - 10t\overset{^}{j}\right)dt

= t\overset{^}{i} + \sqrt{3}t\overset{^}{j} - 10\dfrac{t2}{2}\overset{^}{j}

= t\overset{^}{i} + \sqrt{3}t\overset{^}{j} - 5t^{2}\overset{^}{j}

r = t\overset{^}{i} + \left(\sqrt{3}t - 5t^{2}\right)\overset{^}{j}

or x\overset{^}{i} + y\overset{^}{j} = t\overset{^}{i} + \left(\sqrt{3}t - 5t^{2}\right)\overset{^}{j}

On comparing, x=tx = t & & & &...(i)

and y=3t−5t2y = \sqrt{3}t - 5t^{2} & & & &...(ii)

From Eqs (i) and (ii), we get

y=3x−5x2y = \sqrt{3}x - 5x^{2}

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