A particle is projected with a velocity of 30m/s, at an angle of tan∆°=3/4. After 1 second, the particle is moving at an angle ∆ to the horizontal, where tan∆ is given by(g=10m/s).

Question

Arun · Answer

Tanx = 3/4
Sinx = 3/5
Cosx = 4/5

Along vertical after 1s
v = u + at
vSinα = 30Sinx - g
= 30*3/5 - 10
= 8 m/s —-(1)

Horizontal component of velocity will remain constant throughout the journey.
∴ vCosα = uCosx
vCosα = 30*4/5 = 24 —-(2)

Divide equation (1) by (2)
vSinα / vCos α = 8/24 = 1/3
Tan α = 1/3

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A particle is projected with a velocity of 30m/s, at an angle of tan∆°=3/4. After 1 second, the particle is moving at an angle ∆ to the horizontal, where tan∆ is given by(g=10m/s).

A particle is projected with a velocity of 30m/s, at an angle of tan∆°=3/4. After 1 second, the particle is moving at an angle ∆ to the horizontal, where tan∆ is given by(g=10m/s).

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A particle is projected with a velocity of 30m/s, at an angle of tan∆°=3/4. After 1 second, the particle is moving at an angle ∆ to the horizontal, where tan∆ is given by(g=10m/s).

A particle is projected with a velocity of 30m/s, at an angle of tan∆°=3/4. After 1 second, the particle is moving at an angle ∆ to the horizontal, where tan∆ is given by(g=10m/s).

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