A particle is projected with speed 10m/s perpendicular to the inclined planet angle 60 degree with the horizontal. Therange of the projectile on the incline is
Answers
horizontal range of particle on inclined plane is 40√3 m
A particle is projected with speed 10m/s perpendicular to the inclined plane at an angle 60° with horizontal.
so, angle between particle and horizontal, α = 90° + 60° = 150° and inclination angle , β = 60°
initial velocity of particle, u = 10m/s
horizontal range = 2u²sin(α -β)cosα/gcos²β
= 2(10)²sin(150° - 60°)cos150°/gcos²60°
= 200 × sin(180° + 30°) (-√3/2)/10 × (1/4)
= 20 × (-√3) × 2
= -40√3 m [ negative sign shows that displacement of particle is in downward direction ]
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A particle is projected with speed 10m/s perpendicular to the inclined plane at an angle 60° with horizontal.
so, angle between particle and horizontal, α = 90° + 60° = 150° and inclination angle , β = 60°
initial velocity of particle, u = 10m/s
horizontal range = 2u²sin(α -β)cosα/gcos²β
= 2(10)²sin(150° - 60°)cos150°/gcos²60°
= 200 × sin(180° + 30°) (-√3/2)/10 × (1/4)
= 20 × (-√3) × 2
= -40√3 m [ negative sign shows that displacement of particle is in downward direction ]