Question:

A ball whose kinetic energy is $E$, is projected at an angle of $45^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

Updated On: Jul 5, 2022
  • $E$
  • $E / \sqrt{2}$
  • $E / 2$
  • zero
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The Correct Option is C

Solution and Explanation

At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left which is $u_{x}=u \cos \theta$ Given $\theta=45^{\circ}$ $\therefore u_{x}=u \cos 45^{\circ}=\frac{u}{\sqrt{2}}$ Hence, at the highest point kinetic energy $E^{\prime}=\frac{1}{2} m u_{x}^{2} =\frac{1}{2} m\left(\frac{u}{\sqrt{2}}\right)^{2}$ $=\frac{1}{2} m\left(\frac{u^{2}}{2}\right)$ $=\frac{E}{2} \left(\because \frac{1}{2} m u^{2}=E\right)$
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Concepts Used:

Kinetic energy

Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.