Question

a body is projected with kinetic energy k at angle 60 degree with horizontal its kinetic energy at the highest point of its trajectory will be​

Answer 1

Answer:

K.E' = K.E/4

Explanation:

K.E = ½ mv²

At the highest point, vertical component of velocity becomes zero and only the horizontal component is left that is given by

ux = u cosθ  

θ = 60°, so cos60° = 1/2

ux = u/2

K.E at the highest point =  ½ m (u/2)² = K.E/4

Answer 2

Kinetic energy at highest point is K/4.

Given:

• Body thrown at an angle of 60° with horizontal.

• Initial kinetic energy is K

To find:

Kinetic energy at the highest point of the trajectory?

Calculation:

Let the velocity of projection be u :

So, we can say :

$K = \dfrac{1}{2} m {u}^{2}$

Now, we know that:

• At highest point of a projectile, the net velocity of the particle will be $u\cos(\theta)$ , where $\theta$ is the angle of projection.

• So, u_(h) = u × cos(60°) = u/2.

Now, Kinetic energy will be :

$KE= \dfrac{1}{2} m {(u_{h})}^{2}$

$\implies KE= \dfrac{1}{2} m {( \dfrac{u}{2} )}^{2}$

$\implies KE= \dfrac{\dfrac{1}{2} m {u}^{2} }{4}$

$\implies KE= \dfrac{K}{4}$

So, kinetic energy at highest point will be K/4.