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 out the time at which the rocket will reach its max, to the nearest 100th of a second.

y=-16x^2+174x+78
equation

Answer 1

551.06feet

Explanation:

Given the equation

y=-16x^2+174x+78

At maximum height, dy/dx = 0

dy/dx = -32x + 174

0 = -32x+174

32x = 174

x = 174/32

x = 5.4375

GEt the maximum height

y=-16x^2+174x+78

y=-16(5.4375)^2+174(5.4375)+78

y = -473.0625 + 946.125 + 78

y = 551.0625

Hence the maximum height is 551.06feet to the nearest hundredth