# Question #91249

Jan 27, 2017

Use this equation of motion to solve problems in which you do not know the final velocity

$\Delta y = {v}_{o} \Delta t + \frac{1}{2} a {\left(\Delta t\right)}^{2}$

I'll show you how below.

#### Explanation:

The equation above has been derived especially for the purpose of solving a problem in which you do not know the final velocity of the object.

$\Delta y$ is the distance the object will rise or fall during a given time interval.

${v}_{o}$ is the starting velocity of the object

$\Delta t$ is the time interval beginning from when the velocity was ${v}_{o}$. (It is zero in your problem.)

$a$ is the (constant) acceleration of the object during the period $\Delta t$. (It is -10 in your problem. The - sign means it acts downward.)

The first term in the equation tells us how far the object would travel in time $\Delta t$ if it were to remain at velocity ${v}_{o}$ and not accelerate.

The second term tells us the additional distance it travels due to the acceleration. (It is the complicated part!)

So, since in your problem, ${v}_{o} = 0$, all we do is insert the time (1.9 s) and solve for $\Delta y$

$\Delta y = 0 - \frac{1}{2} \left(10\right) {\left(1.9\right)}^{2} = - 18.1 m$

The negative sign means the object is travelling downward.

In the first 5 seconds, it falls

$\Delta y = 0 - \frac{1}{2} \left(10\right) {\left(5\right)}^{2} = - 125 m$