Question

A man stands on the roof of a 10.0 m -tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 34.0 degrees above the horizontal. You can ignore air resistance.

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

&

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

Answer 1

I hope someone can help me with this problem..I'm losing my mind :(

Answer 2

This will take a bit of work! Using the formula\[Eqn\: 1\rightarrow y(t)=y_0+v_{y0}sin(\theta)t_1 + \frac{1}{2}at_1\text{}^2\]you will be able to find the time at which the rock strikes the ground (by setting y(t) = 0). You will be able to find the velocity at which the rock strikes the ground by using:\[Eqn\: 2 \rightarrow y'(t_1)=v_{y0}sin(\theta) + at_1\]The acceleration for these is the acceleration due to gravity: -9.81 ms\(^{-1}\)

For part 2, use \[Eqn \: 3 \rightarrow x(t) = x_0 + v_{x0}cos(\theta)t_1 + \frac{1}{2}at_1 \text{}^2\]Recall that the rock won't accelerate in the x-direction after you throw it, so a=0 here.

Answer 3

So wait the answer is zero? I'm a tad bit confused

Answer 4

Which answer do you think is zero?

Answer 5

The last one...?

Answer 6

Only the acceleration is zero, so you will end up with \[x(t_1) = x_0 + v_{x0}cos(\theta)t_1\]

Answer 7

Oh ok..

Answer 8

Zarathustra... are you still typing a message?

Answer 9

I think that Eqn2 and Eqn3 need to be different

\[Eqn2 -> y'(t) = v_0\sin(\theta) - \gt\]
\[Eqn3-> x'(t) = v_ocos(\theta)\]

Since the minus sign in the y eqn is already taken into account with g, and there is no acceleration in the x direction. Note the first equation gives you the y velocity and the second gives you the x velocity. Then you need to add them like vectors to get the magnitude.

\[|v| = \sqrt{v_y ^2 + v_x ^2}\]

For the horizontal distance, just throw the time you calculate in the first part into

\[x(t) = x_0 + v_0\cos(\theta)t\]

and that'll tell you how far it goes.

Answer 10

eh, response got screwed up
\[v_y(t) = v_o \sin(\theta)t-g t\]
\[v_x(t) = v_ocos(\theta)\]

for eqn1 and eqn2

Answer 11

and come to think of it, you could use the + sign on the g part, but you would have to change the orientation of your coordinate system. i.e. change y_o. I centered it on the ground.

Answer 12

Yeah, I was only considering the velocity in the y-direction. Go with zarathustra.