I encountered this while doing a very basic physics question (so I hope you don't mind if I post it, the problem lies within my math anyways): A stunt vehicle leaves an inclined slope of 28 degrees with a speed of 35 m/s at a height of 52 m above ground level. Air resistance is negligible.
What is the vehicle's time of flight?
$D = \frac 1 2 at^2+V$ (initial vertical velocity) $t$
is the formula for displacement out of which we can get a quadratic equation to solve for $t$ (time)
$0 = 5\,t^2+\sin(28)\,35\,t + 52$
Now when I plug that into the quadratic formula it gives me a negative under the square root and we are definitely not working with complex numbers. How do I solve for $t$?
Thanks,
John.
Best Answer
I think you need $-5$ where you have $5$.
(And "negative square root" isn't exactly the right term. It would be a square root of a negative number. What's negative is not the square root, but something else.)