Studying for a Physics exam and I'm going back through the homework and redoing the problems. I've come to a problem and am a little confused on the equations.

The problem is:

With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 46 m? (b) How long will it be in the air?

For part a, I used the equation: $v^2=v_{0}^2+2a(x-x_{0})$ where a = -9.8 $m/s^2$ (since acceleration due to gravity is in the negative direction when the ball is being thrown up) but the result I get is a negative number, and you can't take the square root of a negative number. I know there's probably some logic I'm missing here, but I don't know what it is - could someone break this down for me?


For part $a$, you need to solve for $v_0$. At maximum height, $v = 0$. So $0 = v^2 + 2(-9.8)(46)$, or $v_0 = \sqrt{2(9.8)(46)} \approx 30$ m/s.

  • $\begingroup$ Ah, that makes sense - I was trying to solve for $v$ instead of $v_{0}$, silly mistake. $\endgroup$ – secondubly Feb 24 '15 at 17:36

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