Question

Ans my this ques plzz​

Answer 1

Let the initial velocity of the stone be u at an angle of θ with the horizontal.

Since the stone reaches a maximum height of 2h,

02−u2sin2θ=2(−g)(2h)

⟹sinθ=2gh−−√u

Let the stone hit the bird flying with velocity v at time t.

Thus h=usinθt−12gt2

⟹h=2gh−−√t−12gt2

⟹t=(4±2√)hg−−√

The distance traveled by bird from initial position to final position=AB in time t2=(4+2√)hg−−√

AB=ucosθt2=v(t2−t1)

Thus ucosθv=t2−t1t2=22√+1

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Answer 2

Answer:

Don't know ur answer...but what will we do now..?...any other alternate other than this ...? ....no social sites please ...but any other study app..reply on ur status..