# Question b43b8

Sep 29, 2015

I found $28 \frac{m}{s}$ directed downwards.(but check my maths)

#### Explanation:

When the object comes back (passing through the releasing point again) it'll have a velocity of $13.3 \frac{m}{s}$ but directed downwards (neglecting anty air resistance)!
So we have: Sep 29, 2015

$v = - \text{27.5 m/s}$

#### Explanation:

So, you know that an object is being thrown straight upwards from an unknown height $h$ above the ground with an initial velocity of $\text{13.3 m/s}$.

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

${\underbrace{{v}_{\text{top}}^{2}}}_{\textcolor{b l u e}{= 0}} = {v}_{0}^{2} - 2 \cdot g \cdot {h}_{1}$

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 $\text{29.6 m}$ below the point of its release.

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

$\text{9.03 m" + "29.6 m" = "38.63 m}$

So, if the object is free falling for a distance of $\text{38.63 m}$, then you can say that

${v}_{\text{x"^2 = underbrace(v_"top"^2)_(color(blue)(=0)) + 2 * g * "38.63 m}}$

The velocity of the object will thus be

${v}_{\text{x" = sqrt( (2 * 38.63"m" * 9.8"ms"^(-2))) = "27.5 m/s}}$

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")#