Finding t in a Projectile Motion Equation

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To solve for t in the projectile motion equation h = vt - 1/2gt^2, one must recognize it as a quadratic equation. The goal is to rearrange the equation to isolate t on one side. This involves manipulating the equation to form ax^2 + bx + c = 0, where the solutions for t can be found using the quadratic formula. In physics, only the positive root is relevant for time. The final expression derived for t is t = (v - sqrt(v^2 - 2gh)) / g, using the provided values of h, g, and v.
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I am trying to work out the problem for t in the equation. h=vt-1/2gt^2. I have the given for h,v, and g. How do I get everything except t on one side of the equation to solve for t.
 
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You learned this in Algebra II. (Substitute x for t).
 
Think back to algebra, that is where you should have learned to solve for "t" when you have "t" to the second power and first power. The buzz word for what you have there is a "quadratic equation"
 
dooner said:
I am trying to work out the problem for t in the equation. h=vt-1/2gt^2. I have the given for h,v, and g. How do I get everything except t on one side of the equation to solve for t.
Note that one has terms of 1, t, t2, which tells one that this is a quadratic equation, so one can solve for the roots of the quadratic equation.

Remember, for ax2 + bx + c = 0, there are two unqiue solutions. However, in physics, one would be the proper solution, for example, the one for which t is positive.

There would be a lot of manipulation to get t or x on one side, and the answers on the other side, but it is doable.
 
I ended up with t=(v-sqrt(v2-2*g*h)/g to get the answer I needed for the given info h=100,g=9.81, and v=50. Thanks
 
wrong v2 is actually v^2
 
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