Question:

A ball thrown by one player reaches the other in $ 2 s $ . The maximum height attained by the ball above the point of projection will be $ (g = 10\, m/s^{2}) $

Updated On: Jun 20, 2022
  • $ 2.5 \,m $
  • $ 5 \,m $
  • $ 7.5 \,m $
  • $ 10 \,m $
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The Correct Option is B

Solution and Explanation

Since the ball reaches from one player to another in $2\,s$, so the time period of the flite, $T = 2\, s$
$\Rightarrow \frac{2u\, sin\,\theta}{g}=2s$
Here, $u$ is the initial velocity and $\theta$ is the angle of projection.
$\Rightarrow u\,sin\, \theta=g \ldots\left(i\right)$
Now, we know that the maximum height of the projection
$H=\frac{u^{2}\,sin^{2}\,\theta}{2g}$
or $H=\frac{\left(u\,sin\,\theta\right)^{2}}{2g}$
On putting the value of $u\, sin\,\theta$ from E $\left(i\right)$, we have
$H=\frac{g^{2}}{2g}=\frac{g}{2}$
or $H=\frac{g}{2}=\frac{10}{2}\,m$
or $H=5\,m$
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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