Question

A ball is projected with kinetic energy Eat an angle of 45° with the horizontal. At the highest point, during its
flight, its kinetic energy will be
​

Answer 1

Answer:

The kinetic energy of the particle will be 1 / 2 \mathrm{E}.1/2E.

Explanation:

We have the values given as under

Let us assume the initial velocity of the particle be “v”

Kinetic energy = 1 / 2 m \mathrm{V}^{2}=E-arrow 11/2mV2=E−arrow1

Particle velocity at the highest point it reaches is = v cos 45

The particle’s kinetic energy at highest point is = 1 / 2 m(v \cos 45)^{2}1/2m(vcos45)2

=1 / 2(\mathrm{mv}^{2})(1 / 2)=1/2(mv2)(1/2) {from equation 1 we can conclude}

=1 / 2 \mathrm{E}=1/2E

Answer 2

Answer: E/2

Explanation:

$\alpha = 45^{0}$ is the angle made by the ball with the horizontal

At the highest point, Horizontal velocity is only velocity = $vcos\alpha$ = $vcos45^{0}$

So, velocity = $v/\sqrt{2}$

Kinetic Energy, E = $(1/2)mv^{2}$

It means, E changes with $v^{2}$

So, kinetic energy at heighest point = $(1/2)m(v/\sqrt{2})^{2} = mv^{2}/4$ = E/2

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