2 Dimensional Question (rock projected from edge of top of building)

  • Thread starter Thread starter DLPhysics
  • Start date Start date
  • Tags Tags
    Building Edge
AI Thread Summary
A rock is projected from a building at an initial velocity of 18.4 m/s at a 35° angle, landing 96 m away horizontally. The problem requires calculating the building's height, the vertical component of the rock's velocity upon impact, and the overall magnitude of that velocity. The time of flight is determined to be approximately 6.37 seconds, with the horizontal velocity calculated as 15.07 m/s. Confusion arises regarding the initial vertical velocity, which should be derived using trigonometric functions, and the correct application of kinematic equations for both vertical and horizontal motions. Clarification on these calculations is needed to solve the problem accurately.
DLPhysics
Messages
11
Reaction score
0

Homework Statement


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)

Homework 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)

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
 
Physics news on Phys.org
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.
 
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.
 
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

= 0 + 4.9(6.369258724)*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.
 
Thanks for the help! I greatly appreciate it!

Damion
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top