# Homework Help: 2 Dimensional Question (rock projected from edge of top of building)

1. Sep 27, 2009

### DLPhysics

1. The problem statement, all variables and given/known data
A rock is projected from the edge of the top of
a building with an initial velocity of 18.4 m/s
at an angle of 35◦ above the horizontal. The
rock strikes the ground a horizontal distance
of 96 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.

The problem now asks for the height of the building, the vertical component of the rock's velocity when it strikes the ground, and the magnitude of the rock's velocity when it strikes the ground.

time = 6.369258724 sec
velocity in the x = 15.07239761 m/sec
acceleration = 9.8 m/sec*2
initial velocity in the y = 0 m/sec (don't know if that's right)

2. Relevant equations
height = (initial velocity in the y)(time) + 1/2(acceleration)(time)*2
time = sqrt(2 * height/acc.)
final velocity in the y*2 = initial velocity in the y*2 + 2(acceleration)(height)
final velocity in the y = initial velocity in y + (acceleration)(time)

3. The attempt at a solution
I attempted to use trigonometry to find the vertical component of the rock's velocity, but that did not work, but it did work for the horizontal component, and when finding the final velocity in the y, I used the third equation listed, but that did not work either. And, I don't quite understand what the question means by "the magnitude of the rock's velocity when it strikes the ground". I'm not sure whether the initial velocity in the y is 0, 18.4, or some other number, which is probably why I'm having trouble getting the height of the building.

Can anyone help?
Damion

2. Sep 27, 2009

### Delphi51

Sketch a vector 35 degrees above horizontal and mark it 18.4 m/s. Drop a vertical line from the end and make a horizontal line from the beginning so you have a right triangle. Mark the horizontal part Vx and the vertical part Vy. Note that sin (35) = Vy/18.4. Multiply both sides by 18.4 to get Vy = 18.4*sin(35).
Similarly use cosine to find the Vx.

To organize the trajectory motion problem, I suggest you make two headings: horizontal and vertical. In each case ask yourself if the motion is accelerated or not and then write the appropriate formulas under the headings. Fill in the numbers and see what you can find! If you show your work here, we can check it and make suggestions.

3. Sep 27, 2009

### DLPhysics

Vy=10.55380643 m/sec (I did this before, but the HW said that it wasn't the vertical component of the velocity of the rock when it struck the ground

a = 9.8 (positive because the rock is falling)

Viy = ?

Vf = Vi + at

Vf = 0 + (-9.8)(6.369258724)

Vf = -62.418973549

h = Viy (time) + 1/2(acc.)(time)*2
= 0 + 4.9(6.369258724)*2
= 4.9 (40.56745669)
= 198.7805378 (HW said this was wrong)

That's all I have so far, and I'm STILL confused on how to find the magnitude of the rock's velocity.

4. Sep 27, 2009

### Delphi51

Horizontal:
d = vt
96 = 15.07 t
t = 6.37 s. This is the time of flight to ground strike.

Vertical:
d = Vi*t + .5*a*t^2

is not quite right. The Vi in the vertical direction is 10.55 as you said earlier.
Also, you need a minus sign on the 4.9 because the 10.55 is upward and the acceleration is downward.

5. Sep 27, 2009

### DLPhysics

Thanks for the help! I greatly appreciate it!

Damion