Question

the ratio of the potential and kinetic energies of a body projected at an angle 45° with the horizontal when it is at the highest point​

Answer 1

The ratio of the potential and kinetic energies of a body projected at an angle 45° with the horizontal when it is at the highest point is 1.

Explanation:

The angle of projection of the body projected, θ = 45°

Let the velocity of projection be denoted as “u” m/s.

At the highest point, uy (in vertical direction) = 0 & ux (in horizontal direction) = u cosθ

The maximum height attained by the object at the highest point,

h = [u² sin²θ]/2g ….. (i)

Now, at the highest point,

The Potential Energy of the projected body,  

P.E. = mgh = m*g*[u² sin²45°]/2g …… [substituting value of h from (i)] …….. (ii)

And,  

The Kinetic Energy of the projected body,  

K.E. = ½ * m * (ux)² = ½ * m * (u cos45°)² …… (iii)

Thus,  

The ratio of the potential and kinetic energies of a body projected is,

= P.E. / K.E.

= [m*g*{u² sin² 45°}/2g] / [½ * m * (u cos45°)²]

cancelling the similar terms

= sin²45°/ cos²45°

= tan²45°

= 1

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Also view:

If the kinetic energy of an oblique projectile at its  maximum height is half of its initial kinetic energy  then the angle of throw with the vertical is

(1) 30°

(2) 37°

(3) 45°

(4) 60°

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A body is projected such that its kinetic energy at the top is 3/4 th  of its initial kinetic energy.What is the initial angle of projection of the projectile with the horizontal ?

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