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

In summary: Sketch did a great job summarizing the content of the conversation. In summary, a rock is projected from the edge of the top of a building and has an initial velocity of 18.4 m/s. It is then dropped from a height of 35 degrees above the horizontal and strikes the ground a horizontal distance of 96 meters from the base of the building. The rock's acceleration is 9.8 m/s2, and the ground is level and the side of the building is vertical. The problem 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. DamionSketch was able to find the height
  • #1
DLPhysics
11
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
 
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  • #2
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
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
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.
 
  • #5
Thanks for the help! I greatly appreciate it!

Damion
 

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

1. How do you calculate the velocity of the rock projected from the edge of a building?

The velocity of the rock can be calculated using the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the building.

2. What factors affect the trajectory of the rock projected from the building?

The trajectory of the rock is affected by factors such as the initial velocity, air resistance, and the angle at which it is projected. Other factors can include wind speed and direction, as well as the shape and weight of the rock.

3. Can the shape of the building affect the distance the rock travels?

Yes, the shape of the building can affect the distance the rock travels. If the building has a slanted or curved surface, it can alter the trajectory of the rock and potentially increase or decrease the distance it travels.

4. Is it possible to predict where the rock will land?

With the right calculations and understanding of the factors involved, it is possible to predict the general area where the rock will land. However, other unpredictable factors such as wind gusts or imperfections in the rock's shape can affect the final landing spot.

5. How does the height of the building impact the rock's velocity?

The height of the building directly impacts the rock's velocity as it is a key component in the velocity formula. The higher the building, the greater the velocity of the rock will be when it is projected from the edge.

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