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
Answers
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
-----------------------------------------------------------------------------------
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°
https://brainly.in/question/10662690
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 ?
https://brainly.in/question/489269