# A projectile moves in a path following the formula y = 16x^2 + 80. At what point is the velocity of the projectile parallel to the ground.

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### 1 Answer

This is a question involving math rather than physics.

The path of the projectile is given by the equation y = 16x^2 + 80. The point at which the velocity of the projectile is parallel to the ground has to be determined.

Treating the ground as the x-axis, the velocity at a point is parallel to the ground when the tangent to the path at the point is is parallel to the x-axis. A line parallel to the x-axis has a slope equal to 0.

For a function y = f(x), the slope of of the line at any value of x is given by f'(x). Here, y = 16x^2 + 80

=> y'= 32x

Solving y' = 0 gives 32x = 0 or x = 0

The value of y at x = 0 is y = 16*0^2 + 80 = 80

**The projectile has a velocity parallel to the ground at the point (0, 80)**