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A projectile is launched horizontally (i.e., parallel to the inclined surface) from an inclined plane tilted at 45 deg to the horizontal, with initial speed 50 m/s. Taking g = 10 m/s², find the range measured along the inclined surface.

1. 500 m
2. 500*sqrt(2) m
3. 200*sqrt(2) m
4. none of these

Correct answer: 500*sqrt(2) m

Solution

The projectile is fired horizontally from an incline at 45 deg. 'Horizontally' here means in the horizontal direction (not along the incline). So the velocity is horizontal (magnitude 50 m/s). The incline is at 45 deg, so resolving: u_along_incline = 50*cos(45) = 25*sqrt(2) m/s, u_perp_incline = 50*sin(45) = 25*sqrt(2) m/s (away from surface). For the projectile to return to the incline, the perpendicular displacement = 0. a_along = -g*sin(45) = -5*sqrt(2), a_perp = -g*cos(45) = -5*sqrt(2). Time of flight: 0 = 25*sqrt(2)*t - (1/2)*5*sqrt(2)*t² => t = 2*25*sqrt(2)/(5*sqrt(2)) = 10 s. Range along incline = 25*sqrt(2)*10 - (1/2)*5*sqrt(2)*100 = 250*sqrt(2) - 250*sqrt(2) = 0. That's wrong. Let me redo: if fired horizontally from the TOP of the incline, the initial velocity has no component along incline vertically but is purely horizontal.

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