Question

A particle is projected at an angle
from the horizontal with kinetic
energy T. The kinetic energy at the
highest point is

​

Answer 1

Answer:

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: QuEsTiOn :- \: \star}}}}}$

A particle is projected at an angle from the horizontal with kinetic energy T. The kinetic energy at the highest point is

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

${\mathfrak{Kinetic \ energy \ at \ highest \ point \ (K) = Tcos^2 \theta}}$

$\Large{\underline{\underline{\bf{GiVen:-}}}}$

A particle is projected at an angle from the horizontal with kinetic energy T.

$\Large{\underline{\underline{\bf{Find:-}}}}$

What's the Kinetic energy at the highest point ?

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

So as per the Question ,

We ,

Let the angle of projection with horizontal be 'θ' and initial velocity of projectile be 'u' and mass of particle be 'm'

So, Kinetic energy at the point of projection will be:

$\rm T = \dfrac{1}{2} mu^2$

Velocity at highest point of projectile is:

$\rm v = u \: cos \theta$

Kinetic energy at highest point of projectile:

$\rm K = \dfrac{1}{2}m {v}^{2} \\ \\ \rm K = \dfrac{1}{2}m {(u \: cos \theta)}^{2} \\ \\ \rm K = (\dfrac{1}{2}m u ^{2} ) \: cos^{2} \theta \\ \\ \rm K = T \: cos^{2} \theta$

$\therefore\underline{\boxed{\textsf{Kinetic Energy is = {\textbf{Tcos²θ}}}}} \qquad\qquad \bigg\lgroup\bold{Required \ answer} \bigg\rgroup$