The equation of a parabola is y = ax2 + bx + c, where a, b, and c are constants.

Question

The equation of a parabola is y = ax2 + bx + c, where a, b, and c are constants. The x- and y-coordinates of a projectile launched from the origin as a function of time are given by x = vx0t and
y = vy0t -
1
2
gt2,
where vx0 and vy0 are the components of the initial velocity.
(a) Eliminate t from these two equations and show that the path of a projectile is a parabola and has the form y = ax + bx2. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) What are the values of a, b, and c for the projectile? (Use any variable or symbol stated above as necessary.)

Explanation / Answer

Given y = ax^2 +bx + c x = vx0 t , y = vy0t -1/2 gt^2 a) x = (vi cos?) t     t = x/(vi cos?) ----------(1)    y = ( vi sin?)t -1/2 gt^2 substitute the value of (1) in Y component y = (vi sin?) * x/(vi cos?) -1/2 g (x/(vi cos?))^2      = (tan?) x +(-g/2 vi^2 cos^2?) x^2 This equation is in the form of y = ax +bx^2------(2) --------------------------------------------------------------------------- b)In equation (2), the values of a = tan? and b = (-g/2 vi^2 cos^2?)

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