Question

A man stands on the roof of a 19.0 m-tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 34.0 above the horizontal. You can ignore air resistance.

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

Answer 1

There are a couple ways to do this; do you prefer kinematics or conservation of energy methods?

Answer 2

They both reduce to the same equation in the end:
\[(v_f)^2-(v_o)^2 = 2a(y_f-y_o)\]

Answer 3

kinematics

Answer 4

I still need help

Answer 5

Ok. Do you recognize the formula I put there?

Answer 6

It can be derived either by combining the first two fundamental kinematics equations,
\[x_f=x_o+v_o(t)+\frac{1}{2}a(t)^2 \space \space and \space \space v_f=a(t)\]
or by using conservation of energy, i.e. kinetic + potential before = kinetic + potential after.
Either way you can solve for final velocity with
\[(v_f)^2−(v_o)^2=2a(y_f−y_o) \rightarrow v_f=\sqrt{(v_o)^2+2a(y_f−y_o)}\]

Answer 7

And since it only wants the magnitude of the velocity, you don't have to worry about the vectors (except in the positive vs. negative sense).

Answer 8

did you get 36.2 as the answer?

Answer 9

(sorry for the delay, was working on a different problem)
(I haven't worked it out yet, so let me go grab my calculator..)

Answer 10

I got 35.68, so you probably just have some round-off error in there somewhere.