Question

The path of a projectile launched from a 20-ft-tall tower is modeled by the equation y = -5x2 + 40x + 20. What is the maximum height, in meters

reached by the projectile?

Answer 1

30.49 m

Explanation:

To obtain the maximum height, we solve for the value x when dy/dx = 0.

Since, y = -5x² + 40x + 20

dy/dx = d[-5x² + 40x + 20]/dx

dy/dx = -10x + 40

Since dy/dx = 0,

-10x + 40 = 0

-10x = -40

x = -40/-10

x = 4

Substituting x = 4 into the equation for y, we have

y = -5x² + 40x + 20

y = -5(4)² + 40(4) + 20

y = -5(16) + 160 + 20

y = -80 + 160 + 20

y = 80 + 20

y = 100 ft

Since y is in feet, we convert to meters.

Since 1 m = 3.28 ft, 100 ft = 100 ft × 1 m/3.28 ft = 30.49 m

So, the maximum height, in meters reached by the projectile is 30.49 m