Quadratic equations

Question

The path of a projectile launched from a 16 foot tall tower is modeled by the equation y=16x^2×64x+16. When will it hit the ground?

Answer 1

I think there's a mistake in the equation you have listed. Projectile motion is typically modeled as y = -16x² . . . (negative x²) otherwise you get a parabola that goes down to a minimum instead of up to a maximum like a projectile would go. So. I'm going to assume this is

y = -16x² + 64x + 16 where x represents time in seconds and y represents height in feet.

The equation would then model this situation:

To find the place where the function crosses the x-axis, we would make y = 0 and solve:

0 = -16x² + 64x + 16

Using the quadratic formula: x = [-b ± √(b² - 4ac)]/(2a) = [-64 ± √(64² - 4•(-16)•16)]/(2•(-16)) = (-64 ± √5120)/-32 or x = -0.24 sec and x = 4.24 sec. We would throw out the negative time answer and choose the x = 4.24 seconds time