Calculating Projectile Motion: Rock Kicked From a Building

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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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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.
 
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
 
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
 
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]
 
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 : )