Physics, asked by kavita12178, 6 months ago

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

Answers

Answered by Intelligentcat
42

\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

Answered by suneetha25
0

Explanation:

If the particle is projected with velocity , then its initial kinetic energy is Velocity of the projectile at the highest point K.E. of the particle at the highest point

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