Question f48ee

Jan 9, 2017

$\text{153 J}$

Explanation:

The question is essentially asking for the height above ground level at which the object will find itself $\text{2 s}$ after being thrown vertically upward.

As you know, an object's gravitational potential energy, $U$, at a heigh $h$ above ground is equal to the work done against gravity to lift the object to that height.

The equation that allows you to calculate an object's gravitational potential energy looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{U = m \cdot g \cdot h}}}$

Here

• $m$ is the mass of the object
• $g$ is the gravitational acceleration, usually given as ${\text{9.81 ms}}^{- 2}$
• $h$ is the height above ground

Now, in order to find the height above ground at which the object finds itself, you can use the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{h = {v}_{0} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}}}}$

Here

• ${v}_{0}$ is the initial velocity of the object
• $t$ is the total time of flight

Plug in your values to find

h = "15 m" color(red)(cancel(color(black)("s"^(-1)))) * 2 color(red)(cancel(color(black)("s"))) - 1/2 * "9.81 m" color(red)(cancel(color(black)("s"^(-2)))) * 2^2 color(red)(cancel(color(black)("s"^2)))

$h = \text{10.38 m}$

So, the object is located $\text{10.38 m}$ above ground $\text{2 s}$ after its launch, which means that its gravitational potential energy will be

U = "1.5 kg" * "9.81 m s"^(-2) * "10.38 m" = color(darkgreen)(ul(color(black)("153 J")))#

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the time of flight.