A ball- whose kinetic energy is E- is projected at an angle of 45- from the horizontal- What will be it-s kinetic energy at the highest point of its flight-

Suppose the ball of mass m be projected with an initial velocity u. E=1/2mu^2 At the highest point, component of velocity in vertical direction v_y =0 Componen ...

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Standard XII

Physics

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A ball, whose...

Question

A ball, whose kinetic energy is E, is projected at an angle of $45^{\circ}$ from the horizontal. What will be it's kinetic energy at the highest point of its flight?

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Solution

Suppose the ball of mass m be projected with an initial velocity u.
$E = \frac{1}{2} m u^{2}$
At the highest point, component of velocity in vertical direction $v_{y} = 0$
Component of velocity in horizontal direction, $v_{x} = u c o s 45^{\circ} = \frac{u}{\sqrt{2}}$
So, kinetic energy of the ball at highest point,
$K . E = \frac{1}{2} m (\frac{u}{\sqrt{2}})^{2} = \frac{(m u^{2})}{4} = \frac{E}{2}$

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