Question

A cricket player throws ball in air at 60degree with horizontal axis with K. E=50j .The k.E of ball at heighest point of its flight will be​

Answer 1

Answer :

Kinetic energy at initial point = 50J

Angle of projection = 60°

We have to find kinetic energy of ball at highest point$.$

★ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

Let body is projected at a velocity of u.

Velocity of object at the highest point of parabolic path (v) = u cosθ

➝ K = 1/2 mv²

➝ K = 1/2 m(u cosθ)²

➝ K = 1/2 mu² cos²θ

➝ K = 50 × cos²60

➝ K = 50 × 1/4

➝ K = 12.5 J

Answer 2

ㅤ$\;\;\underline{\textbf{\textsf{ Given :-}}}$

• $\sf{\theta=60^o}$

• $\sf{K=50\ J}$

ㅤ$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Kinetic energy of ball at highest point

ㅤ$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{ We know that :}}$

At the highest point the ball will have horizontal velocity only.

• $\sf{v=u\cos\theta}$

• $\sf{K=\dfrac{1}{2}\,mu^2}$

Then, kinetic energy of ball at highest point will be:-

$\sf{\dashrightarrow K'=\dfrac{1}{2}\,mv^2}$

$\sf{ \dashrightarrow K'=\dfrac{1}{2}\,m(u\cos\theta)^2}$

$\sf{ \dashrightarrow K'=\dfrac{1}{2}\,mu^2\cos^2\theta}$

$\sf{ \dashrightarrow K'=K\cos^2\theta}$

$\sf{\dashrightarrow K'=50\cos^260^o}$

$\sf{\dashrightarrow K'=50\times\dfrac{1}{4}}$

$\dashrightarrow {\boxed{\frak{\purple{k' = 12.5\;J}}}}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Kinetic energy of ball at highest point will be </p><p>\textbf{ 12. 5J}}}.$

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