Projectile Motion?

Question

A man stands on the roof of a building of height 16.6 m and throws a rock with a velocity of magnitude 31.1 m/s at an angle of 27.3' above the horizontal. You can ignore air resistance.
*The maximum height above the roof reached by the rock: 10.4 m
*Take free fall acceleration to be 9.80 m/s^2

A) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

B) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.



**please show how you got your answer!! THNX!!!**

Answer 1

A) You can get the final vertical velocity in one step or two.

One step: V^2 = Vo^2 + 2*a*y
where V is the final velocity, Vo=0 (at the top of the trajectory), and y = the height it fell from - the top of the trajectory.

Two steps: [Might as well use this because it'll give you time and you'll need time to solve B).]
y = (1/2)*g*t^2
where y is the same as above and you were given g. Then take that t and use it in
V = Vo + g*t
where Vo and g are as above.

The V you get from that is only the final vertical velocity. Add the horizontal velocity, 31.1*cos(angle) m/s. (Form the resultant vector.)

B) The horizontal velocity while all the vertical stuff is going on is 31.1*cos(angle) m/s. Multiply by the time.