Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

time after launch, x in seconds, by the given equation. Using this equation, find the

maximum height reached by the rocket, to the nearest tenth of a foot.

y = -16x^ + 245x + 104

Answer 1

Answer:

1,041.9feet

Step-by-step explanation:

Given the height of the rocket expressed as

y = -16x² + 245x + 104

At maximum height, dy/dx = 0

dy/dx = -32x+245

0 = -32x+245

32x = 245

x = 245/32

x = 7.65625

Get the maximum height

Recall that;

y = -16x² + 245x + 104

Substitute the value of x;

y = -16(7.65625)² + 245(7.65625) + 104

y = -937.890625 + 1,875.78125 + 104

y = 1,041.890625feet

Hcne the maximum height to the nearest foot is 1,041.9feet