Question

one second after projection a stone moves at an angle of 45 two seconds after projection it moves horizontally find the angle of projection and initial velocity

Answer 1

Hey dear FřeïŇðs ☺️☺️❤️




Ur answer is here....

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ANSWER
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X = u cos θ t                 and   Vx = u cos θ
y = u sin θ t - 1/2 g t²     abd   Vy = u sin θ - g t

The horizontal component of speed Vx is always constant.  So the speed is minimum when the vertical component Vy is 0.
=> u sinθ = 2 g 

45° = direction of velocity of the project at  time t = 1 sec.
Then  Tan 45° = 1 = Vy / Vx = (u sinθ - g*1)/(u cosθ)
=> u cosθ = g
=>  tan θ = 2    =>  θ = Tan⁻¹ 2
=>  sinθ = 2/√5  and cosθ = 1/√5     and u = √5 g



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Hope its help uh.......
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__SUMIT TIWARI FROM [UP] AZAMGARH

Answer 2

Answer:

th e answer for this question is tan inverse of 2

and intial velocity is square root og 5 g