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The equation of motion of a projectile is y 12 x- 3/4 . Given that g 10 s2 what is the rangsthe projectile? (the origin is the point of projection, x-axis is horizontal and y-axis is vertical(A) 36m(R) 30 6 m |
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Answer» y=12x-3/4x^2 The simplest way to calculate the range is to put y=0 in the equation of parabolic trajectory. This gives- 12x-3/4x^2=0. Therefore, x(12–3/4x^2)=0. Then, x=0 gives x coordinate of the point of projection and 12=3/4x or x=16 m gives the range of the projectile. We can find range by noting that at x=R/2, dy/dx=0. dy/dx=12–3/2x=0 for x=R/2. Thus, 12–(3/2)R/2=0. Therefore, R=16m. |
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