# Question #b43b8

##### 2 Answers

#### Answer:

I found

#### Explanation:

When the object comes back (passing through the releasing point again) it'll have a velocity of

So we have:

#### Answer:

#### Explanation:

So, you know that an object is being thrown *straight upwards* from an **unknown** height

At **meximum height**, the velocity of the object will be equal to **zero**, which means that you can say

#underbrace(v_"top"^2)_(color(blue)(=0)) = v_0^2 - 2 * g * h_1#

The distance the object travels before reaching *maximum height* will thus be

#h_1 = v_0^2/(2 * g) = (13.3^2 "m"^2color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2))))) = "9.03 m"#

Now, you know that you must determine the object's velocity when it is located **below** the point of its release.

This is equivalent to saying that you must find its velocity after the fell for a total of

#"9.03 m" + "29.6 m" = "38.63 m"#

So, if the object is *free falling* for a distance of

#v_"x"^2 = underbrace(v_"top"^2)_(color(blue)(=0)) + 2 * g * "38.63 m"#

The velocity of the object will thus be

#v_"x" = sqrt( (2 * 38.63"m" * 9.8"ms"^(-2))) = "27.5 m/s"#

If we take the *upwards direction* to be the **positive direction**, then the velocity of the object will be

#v_"x" = color(green)(-"27.5 m/s")#