A particle is projected in the x-y plane with y-axis along vertical. Two second after projection the velocity of the particle makes an angle 45° with the x-axis. Four second after projection, it moves horizontally. Find the velocity of projection.kindly answer this question.....
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Projectile: Initial speed = u , initial angle of projection: Фo.
Velocity makes an angle Ф with horizontal at time t.
Tan Ф = dy/dx = Vy / Vx = (u SinФo - g t ) / (u CosФo)
=> Tan Ф = TanФo - (g SecФo /u) t --- (1)
1. At t = 2 sec., Ф = 45°.
Tan 45° = 1 = Tan Фo - (2 g Sec Фo ) /u --- (2)
2. At t = 4 sec., Ф = 0°.
Tan 0 = 0 = Tan Фo - (4 g Sec Фo) / u --- (3)
=> u = 4 g Cosec Фo --- (4)
Substitute (4) in (2):
Tan Фo = 2 => Cosec Фo = √5/2
u = 2 √5 g
Velocity makes an angle Ф with horizontal at time t.
Tan Ф = dy/dx = Vy / Vx = (u SinФo - g t ) / (u CosФo)
=> Tan Ф = TanФo - (g SecФo /u) t --- (1)
1. At t = 2 sec., Ф = 45°.
Tan 45° = 1 = Tan Фo - (2 g Sec Фo ) /u --- (2)
2. At t = 4 sec., Ф = 0°.
Tan 0 = 0 = Tan Фo - (4 g Sec Фo) / u --- (3)
=> u = 4 g Cosec Фo --- (4)
Substitute (4) in (2):
Tan Фo = 2 => Cosec Фo = √5/2
u = 2 √5 g
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