A model rocket is launched straight upward from the side of a 212-ft cliff. The initial velocity is 86 ft/sec. The height of the rocket h(t) is given by: h(t)=-16t^2+86t+212
where h(t) is measured in feet and t is the time in seconds. Use the quadratic formula to find the times (t) at which the rocket is 320 ft. above the ground. Round your answers to the nearest hundredth of a second.
just solve for t in
-16t^2+86t+212 = 320
To find the times at which the rocket is 320 ft above the ground, we need to solve the equation h(t) = 320.
Given the quadratic equation h(t) = -16t^2 + 86t + 212, we can substitute 320 for h(t):
-16t^2 + 86t + 212 = 320
Rearranging the equation to form a quadratic equation in standard form:
-16t^2 + 86t + 212 - 320 = 0
-16t^2 + 86t - 108 = 0
We now have a quadratic equation in the form of ax^2 + bx + c = 0, where a = -16, b = 86, and c = -108.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
t = (-(86) ± √((86)^2 - 4(-16)(-108))) / (2(-16))
Now, let's calculate:
t = (-86 ± √(7396 - 6912)) / (-32)
t = (-86 ± √(484)) / (-32)
We have two possible solutions:
t = (-86 + √484) / (-32) or t = (-86 - √484) / (-32)
Simplifying further:
t = (-86 + 22) / (-32) or t = (-86 - 22) / (-32)
t = -64 / -32 or t = -108 / -32
t = 2 or t = 3.375
Therefore, the times at which the rocket is 320 ft above the ground are approximately 2 seconds and 3.375 seconds.