Question:

When $a$ ball is thrown up vertically with velocity $v_0$, it reaches a maximum height of $h$. If one wishes to triple the maximum height then the ball should be thrown with velocity

Updated On: Aug 1, 2022
  • $\sqrt{3}v_{0}$
  • $3\,v_0$
  • $9\,v_0$
  • $3/2\,v_0$
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The Correct Option is A

Solution and Explanation

At maximum height velocity is zero. From equation of motion we have $v^{2}=u^{2}-2 g h$ where $v$ is final velocity, $u$ is initial velocity. Since ball reaches maximum height, velocity at the highest point is zero. Therefore, we have $v=0, u=v_{0}$ $0=v_{0}^{2}-2 g h$ $\Rightarrow v_{0}=\sqrt{2 g h}$ when $h'=3 h$ then $v_{0}'=\sqrt{2 g \times 3 h}=\sqrt{3} \sqrt{2 g h}=\sqrt{3} v_{0'}$ If ball has to be thrown to a greater height its initial velocity should be more than the original one.
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Questions Asked in AIIMS exam

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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