A ball whose kinetic energy is E, is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

Question

A ball whose kinetic energy is E, is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be : (A) E (B) E / √2 (C) E / 2 (D) zero

Tardigrade Admin · Accepted Answer

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 θ

Given

θ = 4 5^{\circ}

∴ u_{x} = u cos 4 5^{\circ} = \frac{u}{2}

Hence, at the highest point kinetic energy

E^{'} = \frac{1}{2} m u_{x}^{2} = \frac{1}{2} m (\frac{u}{2})^{2}

= \frac{1}{2} m (\frac{u ^{2}}{2})

= \frac{E}{2} (∵ \frac{1}{2} m u^{2} = E)

Q. A ball whose kinetic energy is $E$ , is projected at an angle of $4 5^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

$E$

$E / 2$

$E /2$

zero

Q. A ball whose kinetic energy is E, is projected at an angle of 45∘ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

E

E/2​

E/2

zero

At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left which is ux​=ucosθ Given θ=45∘ ∴ux​=ucos45∘=2​u​ Hence, at the highest point kinetic energy E′=21​mux2​=21​m(2​u​)2 =21​m(2u2​) =2E​(∵21​mu2=E)

Q. A ball whose kinetic energy is $E$ , is projected at an angle of $4 5^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be :

$E$

$E / 2$

$E /2$