Calculating Projectile Motion: Rock Kicked From a Building

In summary, the rock is kicked off the top of a building and it hits the ground 71.6 meters away from the top.
  • #1
davidatwayne
9
0
kicking rock of a building...

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.
 
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  • #2
davidatwayne said:

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.
19.8m/s is the initial velocity and is given in the horizonatl direction. Ther is no y component of the initial velocity. You've got to use your motion equations in each direction, and note thatt in the x direction, there is no acceleration, since no forces act in that direction during the flight.
 
  • #3
davidatwayne said:

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.


1) initial position and initial velocity
This is a 2-D problem so you have x (horizontal) and y (vertical) components

initial position : [tex]x_i = 0 e_x + 64 e_y[/tex]
initial velocity : [tex]v_i = 19.8 e_x + 0 e_y[/tex]

Gravity works along the y-axis, which you implemented correctly (-9.81)

2) So applying Newton's second law gives us :

[tex]a_i = 0 t^2/2 + 19.8 t[/tex]
[tex]a_y = -9.81 t^2/2 + 0 t + 64[/tex]

Do you understand these equations ?

If so, solve them for t and you will have your answer

marlon
 
  • #4
I am still not getting it...

I tried using the quadratic formula to solve for the time, using

64 (y) = [19.8 (initial velocity) x t] - (1/2)(9.8)t^2

I came up with an answer of 2.02 seconds.

then I used:

x = [v (initial velocity) x time] + (1/2) x (-9.8) x (time squared)

came up with - 19.98.
which is not the correct answer, nor can a change in x be negative
 
  • #5
davidatwayne said:
I tried using the quadratic formula to solve for the time, using

64 (y) = [19.8 (initial velocity) x t] - (1/2)(9.8)t^2
You must take care to treat horizontal and vertical motion separately. What's the y-component of the initial velocity? (Recall that the rock is kicked horizontally.)

Also, take care with the signs. Use + for up and - for down. The full equation for the vertical position is:
[tex]y_f = y_i + v_{i,y} t - (1/2) (9.8) t^2[/tex]
 
  • #6
ok, i think i got it... i was looking for how v for the x direction and v for the y direction were related... thanks for your help guys : )
 

What is the purpose of "Kicking rock of a building"?

The purpose of "Kicking rock of a building" is to determine the structural integrity and stability of a building by kicking a small rock at it. This method is often used in building inspections and can reveal any potential weaknesses or damage.

How is "Kicking rock of a building" performed?

To perform "Kicking rock of a building", a small rock is selected and kicked towards the building at various locations and heights. The impact and resulting vibrations are then observed and analyzed to determine the building's structural soundness.

What are the benefits of using "Kicking rock of a building" in building inspections?

Using "Kicking rock of a building" in building inspections allows for a quick and easy assessment of a building's structural stability. It can also provide valuable information on potential areas of concern that may need further investigation.

Are there any limitations to using "Kicking rock of a building"?

While "Kicking rock of a building" can provide useful information, it should not be the sole method of determining a building's structural integrity. Other factors such as age, construction materials, and previous maintenance should also be taken into consideration.

Is "Kicking rock of a building" a widely accepted method in the scientific community?

While "Kicking rock of a building" may be a commonly used method in building inspections, it is not considered a scientifically rigorous or reliable method. Other non-destructive testing methods, such as ground-penetrating radar, are preferred for more accurate and thorough assessments of a building's structural stability.

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