Question

A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 200t. How many seconds after the launch will the rocket be 350 ft above the ground? Round to the nearest hundredth of a second. (Enter your answers as a comma-separated list.)

Answer 1

Answer:

2.10 s, 10.40 s.

Step-by-step explanation:

We know that the height of the rocket is given by the function:

$h=-16t^2+200t$

We are asked to find the time for which the height of the rocket will be 350 ft. So, for that moment, we know the height but we don't know the time; however, we know that the equation can help us to find the time, doing h=350:

$350=-16t^2+200t$

The last is a quadratic equation, which can be put in the form $at^2+bt+c=0$ and solved applying the formula:

$t=\frac{-b+-\sqrt{b^2-4ac} }{2a}$

So, let's put the equation on the form $at^2+bt+c=0$ adding $16t^2$ and subtracting $200t$ to each side of the equation; the result is:

$16t^2-200t+350=0$

So, we note that a=16, b=-200, and c=350.

Then,

$t_1=\frac{200-\sqrt{200^2-4*16*350} }{2*16}=2.10$

$t_2=\frac{200+\sqrt{200^2-4*16*350} }{2*16}=10.40$

According to the equation, that are the times for which the height will be 350 ft; that is because the rocket is going to ascend and then to fail again to the ground.