Question

A model rocket is launched with an initial velocity of 240 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 240t.
How many seconds after launch will the rocket be 390 ft above the ground? Round to the nearest hundredth of a second.

s (smaller value)
s (larger value)

Answer 1

Answer:

About 1.85 seconds and 13.15 seconds.

Step-by-step explanation:

The height (in feet) of the rocket t seconds after launch is given by the equation:

$h = -16t^2 + 240 t$

And we want to determine how many seconds after launch will be rocket be 390 feet above the ground.

Thus, let h = 390 and solve for t:

$390 = -16t^2 +240t$

Isolate:

$-16t^2 + 240 t - 390 = 0$

Simplify:

$8t^2 - 120t + 195 = 0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}$

In this case, a = 8, b = -120, and c = 195. Hence:

$\displaystyle t = \frac{-(-120)\pm \sqrt{(-120)^2 - 4(8)(195)}}{2(8)}$

Evaluate:

$\displaystyle t = \frac{120\pm\sqrt{8160}}{16}$

Simplify:

$\displaystyle t = \frac{120\pm4\sqrt{510}}{16} = \frac{30\pm\sqrt{510}}{4}$

Thus, our two solutions are:

$\displaystyle t = \frac{30+ \sqrt{510}}{4} \approx 13.15 \text{ or } t = \frac{30-\sqrt{510}}{4} \approx 1.85$

Hence, the rocket will be 390 feet above the ground after about 1.85 seconds and again after about 13.15 seconds.