Recent content by TrueBlood

Figured out a way to do it. First find the max height , then the time it takes to fall from the max height to the ground (so height above original place + 2.4). Then add that time with the time it takes to reach the top. Then simply find range.

If you are given the projectile angle, initial velocity, and height you can find the projectile range using the quadratic equation (solving for t). One thing I don't understand is why the quadratic equation seems to be the only way to solve these problems. For instance, let's say an object is...

The last step, mathematically correct, would be... sqrt[(2h-2Vot)/g] = t and in a free-fall equation, where Vo = 0, then t = sqrt(2h/g) I found out t = Vo*sqrt(2h/g) when you are trying to find the range (r = volt) and using t = sqrt(2h/g) when you don't have t. Thx.

Studying for the MCAT, and trying to figure out how t = sqrt(2h/g) and t = Vo*sqrt(2h/g) is derived from the standard translational motion equations. h = change in h t = change in t thus... h = volt + 1/2g(t)^2 h - volt = 1/2g(t)^2 (2h)/g - (2Vot)/g =t^2 sqrt(2h/g) - sqrt(2Vot)/g =...