A paRticle projected from ground move at 45 degree with horizontal one second after projection and speed is minimum two seconds after projection.The angle of projection of particle is (neglect air resistance)
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Answered by
473
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
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
Answered by
65
We know time of flight is 4 Seconds. As it is 2x time taken to reach 90degrees and speed is minimum when it reaches 90 degrees with vertical.
When speed is minimum VSinΘ = GT
VSinΘ= 2G
TanΘ=Vx/Vy=(USinΘ-gt)/UcosΘ
UcosΘ=G TanΘ=2 Θ=tan-1(2)
Hope it helsp
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