A man stands on the roof of a 16.0 -tall building and throws a rock with a veloc

Question

A man stands on the roof of a 16.0 -tall building and throws a rock with a velocity of magnitude 24.0 at an angle of 30.0 above the horizontal. You can ignore air resistance.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Explanation / Answer

We can separate the velocity into two components. There are an infinite number of ways to do this, but if we choose horizontal and vertical we will have one component (horizontal) that stays fixed and another component that changes as it is acted upon by gravity (vertical) More of the throw is horizontal than vertical cos(30) > sin(30) so V cos(30) = horizontal component and V sin(30) = vertical component Now we find the final vertical component of the velocity. We don't have time we have Vo, g, and change in height. We need Vf The obvious choice is Vf^2 = Vo^2 + 2ad since it doesn't need time. calling up + Vf^2 = [24m/s*sin(30)]^2 +2(-9.8m/s^2)(-16.0m) = 457.92 m^2/s^2 the horizontal component is still 24m/s*cos(30) = 20.78m/s Find the magnitude of the total velocity by adding using Pythagorian Theorem V tot = SQRT(457.92 m^2/s^2 +{20.78m/s}^2) = 29.8 m/s ANSWER magnitude of the velocity of the rock just before it strikes = 29.8 m/s ANSWER

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