Question

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Answer 1

★Given:-

• A ball is thrown with velocity 8m/s
• Angle made with the horizontal = 60°

★To find:-

• Time in which it's velocity will be perpendicular to the direction of initial velocity of projection.

★Solution:-

Let velocity of body after time t is perpendicular upon initial velocity .

Velocity after time 't'

⇒ v = ucosθ i + (usinθ -gt )j

Initial velocity :-

v'= ucosθ i + usinθ j

As v and v' are perpendicular,

⇒ v.v' = 0

Substituting the values,

⇒(ucosθ i + (usinθ -gt) j )(ucosθ i + usinθ j )=0

⇒u²cos²θ +u²sin²θ -usinθ.gt =0

      [cos²θ +sin²θ=1]

⇒u - sinθgt = 0

⇒t = u/sinθg

      =8/sin60° (10)

     =8/√3/2 (10 )

     = 8×2 /10√3

       =1.6/√3 sec

Hence,

Time in which it's velocity will be perpendicular to the direction of initial velocity of projection = 1.6/√3.

Option(A) is correct.

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