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 Example

 Solved Problems

 Difficult Problems

Here, we show you a step-by-step solved example of free fall. This solution was automatically generated by our smart calculator:

A ball is dropped from the highest part of a building that has a height of 20 m. What time does it take to reach the ground?

What do we already know? We know the values for acceleration ($a$), initial velocity ($v_0$), distance ($y$), height ($y_0$) and want to calculate the value of time ($t$)

$a=-9.81\:m/s2,\:\: v_0=0,\:\: y=20\:m,\:\: y_0=0,\:\: t=\:?$

According to the initial data we have about the problem, the following formula would be the most useful to find the unknown ($t$) that we are looking for. We need to solve the equation below for $t$

$y=y_0+v_0t- \left(\frac{1}{2}\right)at^2$

We substitute the data of the problem in the formula and proceed to simplify the equation

$20=0+0t- -9.81\cdot \left(\frac{1}{2}\right)t^2$

Multiply the fraction and term in $9.81\cdot \left(\frac{1}{2}\right)t^2$

$20=0+0t+\frac{9.81\cdot 1}{2}t^2$

Multiply $9.81$ times $1$

$20=0+0t+\frac{9.81}{2}t^2$

Multiply the fraction and term in $9.81\cdot \left(\frac{1}{2}\right)t^2$

$20=0+0t+\frac{9.81}{2}t^2$

Any expression multiplied by $0$ is equal to $0$

$20=0+\frac{9.81}{2}t^2$

$x+0=x$, where $x$ is any expression

$20=\frac{9.81}{2}t^2$

Rearrange the equation

$\frac{9.81}{2}t^2=20$

Multiply both sides of the equation by $2$

$9.81t^2=20\cdot 2$

Multiply $20$ times $2$

$9.81t^2=40$

Multiply both sides of the equation by $2$

$9.81t^2=40$

Divide both sides of the equation by $9.81$

$\frac{9.81t^2}{9.81}=\frac{40}{9.81}$

Simplify the fraction $\frac{9.81t^2}{9.81}$ by $9.81$

$t^2=\frac{40}{9.81}$

Divide both sides of the equation by $9.81$

$t^2=\frac{40}{9.81}$

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{2}$

$\sqrt{t^2}=\sqrt{\frac{40}{9.81}}$

Cancel exponents $2$ and $1$

$t=\sqrt{\frac{40}{9.81}}$

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{2}$

$t=\sqrt{\frac{40}{9.81}}$

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$t=\frac{\sqrt{40}}{\sqrt{9.81}}$

The complete answer is

The time of the ball is $\frac{\sqrt{40}}{\sqrt{9.81}}$ s

 Final answer to the problem