Question

A model rocket is launched with an initial upward velocity of 202 ft/s. The rocket's height h (in feet) after t seconds is given by the following.
h= 202t-16t^2
Find all values of t for which the rocket's height is 82 feet.

Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

Hello!

The answer is:

The rocket will reach a height of 82 feet at:

$t1=0.42s$

and

$t2=12.21s$

Why?

We are asked to find the value(s) of t for which the rocket's height is 82 feet, so, we need to replace the height value (82 feet) into the given quadratic equation, and then, isolate "t" making the equation equal to 0.

So, writing the quadratic formula, we have:

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

The given function is:

$-16x^2+202x=h$

Substituting h equal to 82 feet, we have:

$-16x^2+202x=82$

$-16x^2+202x-82=0$

Where,

$a=-16\\b=202\\c=-82$

Then, substituting it into the quadratic equation we have:

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}=\frac{-202+-\sqrt{202x^{2} -4*-16*-82} }{2*-16}\\\\\frac{-202+-\sqrt{202x^{2} -4*-16*-82} }{2*-16}=\frac{-202+-\sqrt{40804-5248} }{-32}\\\\\frac{-202+-\sqrt{40804-5248} }{-32}=\frac{-202+-\sqrt{35556} }{-32}=\frac{-202+-(188.56) }{-32}\\\\t1=\frac{-202+(188.56) }{-32}=0.42\\\\t2=\frac{-202-(188.56) }{-32}=12.205=12.21$

So, the rocket will reach a height of 82 feet at:

$t1=0.42s$

and

$t2=12.21s$

Have a nice day!