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

About 1.85 seconds and 13.15 seconds.

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 = (-b\pm√(b^2 -4ac))/(2a)`

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

`\displaystyle t = (-(-120)\pm √((-120)^2 - 4(8)(195)))/(2(8))`

Evaluate:

`\displaystyle t = (120\pm√(8160))/(16)`

Simplify:

`\displaystyle t = (120\pm4√(510))/(16) = (30\pm√(510))/(4)`

Thus, our two solutions are:

`\displaystyle t = (30+ √(510))/(4) \approx 13.15 \text{ or } t = (30-√(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.