Question

a projectile is launched at 45 degree to the horizontal with initial kinetic energy E assuming air resistance to be negligible what will be kinetic energy of projectile when it reaches at highest point​

Answer 1

Answer:

The kinetic energy at the highest point is 0.5 times of E (initial kinetic energy).

Explanation:

A projectile is launched at 45 degrees to the horizontal with initial kinetic energy E.

Angle , $\theta$ = 45 degrees

Initial Kinetic energy = E

Let the velocity be v

E = $\frac{1}{2}$ $\times$ m $\times$ $v^{2}$

At the highest point,

       The horizontal component of velocity = v cos $\theta$

       The vertical component of velocity = 0

Therefore kinetic energy at highest point, KE = $\frac{1}{2}$ $\times$ m $\times$ $(v cos\theta)^{2}$

                                                                       KE    = $\frac{1}{2}$ $\times$ m $\times$ $v^{2} \times cos^{2}45$

                                                                         KE    = $\frac{1}{2}$ $\times$ m $\times$ $v^{2}$ $\times$  $\frac{1}{2}$

                                                                          KE    = E $\times$  $\frac{1}{2}$ = 0.5 E