# Quadratic formula with negative square root?

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.

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.)

• oh is that because of the negative acceleration? thanks a lot. However can you explain to me why its a negative acceleration when more than half the time the acceleration of gravity is positive. I dont quite understand. Thanks again. – John Aug 27 '11 at 21:15
• @John: You need to decide which way is up. Gravity $a$ pulls down. The initial velocity $V$ is up. The distance $D$ to ground is down. So if up is positive, then $V$ should be positive and $a$ and $D$ should be negative. If down is positive, then $V$ should be negative and $a$ and $D$ positive. You'll get the same answer either way. – Ilmari Karonen Aug 27 '11 at 21:20
• Thankyou sir, though of it the wrong way :) – John Aug 27 '11 at 21:22

I haven't done any kinematics in years, so I could be wrong.. (in which case I'll be happy to remove this answer!) but I don't think your approach will work. You've set the initial velocity well, but the distance as 52m. We know the vehicle has to travel 52 to meters from the ramp to get to the ground, but you haven't factored in any of the "above 52m" travelling time.

I'll let the initial velocity be denoted as $v_i$. I think what you need to do is calculate how high the vehicle goes up in the air, then calculate from that height, how long it takes to get back to the ground. Namely, let $t_1$ be the time it takes for the vehicle to stop travelling up, and start travelling down, and let $t_2$ be the time from this point to when the cart hits the ground. To compute $t_1$, we have \begin{align*} v_f^2 = v_i^2 + 2ad \end{align*} where $v_f = 0$, so we find \begin{align*} d = -v_i^2/(2a). \end{align*} Notice we are moving up, against gravity, so your value for $a$ should be -10, -9.8,-9.81, or whatever you guys use for gravity. Now that we know how high the cart travels, we can compute the time to get there. We have \begin{align*} v_f = v_i + at_1 \end{align*} and so \begin{align*} t_1 = -v_i/a. \end{align*} To compute $t_2$, we know we traveled $d$ meters above the ramp, so from here, our travel down will be $d + 52$ meters. Hence we have \begin{align*} d + 52 = v_it_2 + \frac{1}{2}at_2^2 \end{align*} (where as we are travelling with gravity, $a$ is now positive) and so \begin{align*} t_2 = \sqrt{\frac{2(d + 52)}{a}}. \end{align*} The cart travels in the air for a time of $t_1 + t_2$.

"How do I solve for $t$?"

In this case, the physical problem obviously has a real, positive solution: the vehicle will end-up on the ground after some time.

So you will solve by fixing the equation. Chase any sign error.

• Simplest answer :) – Mr Pie Jun 18 '18 at 13:10