Question

The vertical height v, in feet, of a snowboarder jumping off of an overhang can be modeled by the function h(t)=15-10t-16t^2, where 15 is the initial height (in feet) of the overhang, -10 is the initial vertical velocity (in feet per second), and t is the time (in seconds). How long does it take the snowboarder to land after jumping off the overhang?"

Answer 1

Step-by-step explanation:

It is given that, the vertical height v, in feet, of a snowboarder jumping off of an overhang can be modeled by the function as :

$h(t)=15-10t-16t^2$

Where

15 is the initial height of the overhang

-10 is the initial vertical velocity

t is in second

We need to find the time taken by the snowboarder to land after jumping off the overhang. At this condition,

h(t) = 0

So, $15-10t-16t^2=0$

On solving above equation using online calculator, we get the value of t = 0.705 seconds. So, the time taken by the snowboarder to land after jumping off the overhang is 0.705 seconds.                                                                                        

Answer 2

I don't know for sure, but I think there might be an error in the way you copied your function. In all of my dealings with these types of problems, I have learned the formula to be h(t) = 15 + 10t - 16t^2. Notice the plus in front of the "10t". This is due to the fact that if he pushes off of the overhang he would have an upward force of 10 feet per second as soon as his feet left the ground. The only thing pulling him back to Earth is gravity, modeled by the "-16t^2". This is derived from a bit of slightly advanced physics involving the gravitational constant, but let's work under my formula for a second...

Either way, we will wind up using the Quadratic Formula (or possibly factoring if the numbers are easy enough to work with). So let's start. 

h(t)= 15 +10t - 16t^2

In order to use the QF or factoring I will need to make h(t)=0. Simply done by:

0= 15 +10t -16t^2

Looking at the numbers, I'd prefer to use the QF so here it is:

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

I know that my answer will need to be positive since you can't have a negative value when dealing with time, so I will eliminate the positive sign from the "+or-" part leaving me with:

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

And I know that a= -16, b= 10, and c=15. So all that's left to do is substitute and solve.

$x= \frac{-10 -\sqrt{-16^{2}-4*-16*15 } }{2*-16}$ 

There's a decent amount of math that would be difficult and sloppy for me to do over the computer, but all you need to do is solve the rest of the equation and you would get your answer.

Exact answer: $\frac{5-4 \sqrt{15} }{8}$ or rounded answer: ≈1.31 seconds.

Hope this helps!
NoThisIsPatrick