Algebra Examples

$h = \frac{1}{2} g t^{2} + v_{0} t$ ; for $t$

Step 1

Rewrite the equation as $\frac{1}{2} \cdot (g t^{2}) + v t = h$ .

$\frac{1}{2} \cdot (g t^{2}) + v t = h$

Step 2

Multiply $\frac{1}{2} (g t^{2})$ .

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Step 2.1

Combine $g$ and $\frac{1}{2}$ .

$\frac{g}{2} t^{2} + v t = h$

Step 2.2

Combine $\frac{g}{2}$ and $t^{2}$ .

$\frac{g t^{2}}{2} + v t = h$

Step 3

Subtract $h$ from both sides of the equation.

$\frac{g t^{2}}{2} + v t - h = 0$

Step 4

Multiply through by the least common denominator $2$ , then simplify.

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Step 4.1

Apply the distributive property.

$2 (\frac{g t^{2}}{2}) + 2 (v t) + 2 (- h) = 0$

Step 4.2

Simplify.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$2 (\frac{g t^{2}}{2}) + 2 (v t) + 2 (- h) = 0$

Step 4.2.1.2

Rewrite the expression.

$g t^{2} + 2 (v t) + 2 (- h) = 0$

Step 4.2.2

Multiply $- 1$ by $2$ .

$g t^{2} + 2 v t - 2 h = 0$

Step 4.3

Move $v$ .

$g t^{2} + 2 t v - 2 h = 0$

Step 5

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6

Substitute the values $a = g$ , $b = 2 v$ , and $c = - 2 h$ into the quadratic formula and solve for $t$ .

$\frac{- 2 v \pm \sqrt{{(2 v)}^{2} - 4 \cdot (g \cdot (- 2 h))}}{2 g}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Add parentheses.

$t = \frac{- 2 v \pm \sqrt{{(2 v)}^{2} - 4 \cdot (g \cdot (- 2 h))}}{2 g}$

Step 7.1.2

Let $u = g \cdot (- 2 h)$ . Substitute $u$ for all occurrences of $g \cdot (- 2 h)$ .

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Step 7.1.2.1

Apply the product rule to $2 v$ .

$t = \frac{- 2 v \pm \sqrt{2^{2} v^{2} - 4 \cdot u}}{2 g}$

Step 7.1.2.2

Raise $2$ to the power of $2$ .

$t = \frac{- 2 v \pm \sqrt{4 v^{2} - 4 u}}{2 g}$

Step 7.1.3

Factor $4$ out of $4 v^{2} - 4 u$ .

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Step 7.1.3.1

Factor $4$ out of $4 v^{2}$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2}) - 4 u}}{2 g}$

Step 7.1.3.2

Factor $4$ out of $- 4 u$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2}) + 4 (- u)}}{2 g}$

Step 7.1.3.3

Factor $4$ out of $4 (v^{2}) + 4 (- u)$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - u)}}{2 g}$

Step 7.1.4

Replace all occurrences of $u$ with $g \cdot (- 2 h)$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - (g \cdot (- 2 h)))}}{2 g}$

Step 7.1.5

Simplify each term.

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Step 7.1.5.1

Rewrite using the commutative property of multiplication.

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} - (- 2 g h))}}{2 g}$

Step 7.1.5.2

Multiply $- 2$ by $- 1$ .

$t = \frac{- 2 v \pm \sqrt{4 (v^{2} + 2 g h)}}{2 g}$

Step 7.1.6

Rewrite $4$ as $2^{2}$ .

$t = \frac{- 2 v \pm \sqrt{2^{2} (v^{2} + 2 g h)}}{2 g}$

Step 7.1.7

Pull terms out from under the radical.

$t = \frac{- 2 v \pm 2 \sqrt{v^{2} + 2 g h}}{2 g}$

Step 7.2

Simplify $\frac{- 2 v \pm 2 \sqrt{v^{2} + 2 g h}}{2 g}$ .

$t = \frac{- v \pm \sqrt{v^{2} + 2 g h}}{g}$

Step 8

The final answer is the combination of both solutions.

$t = - \frac{v - \sqrt{v^{2} + 2 g h}}{g}$

$t = - \frac{v + \sqrt{v^{2} + 2 g h}}{g}$