Question #b43b8

2 Answers
Sep 29, 2015

Answer:

I found #28m/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 #13.3m/s# but directed downwards (neglecting anty air resistance)!
So we have:

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Sep 29, 2015

Answer:

#v = -"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 #"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_"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 #"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

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

So, if the object is free falling for a distance of #"38.63 m"#, then you can say that

#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")#