# Kicking rock of a building

1. Jul 25, 2007

### davidatwayne

kicking rock of a building....

1. The problem statement, all variables and given/known data

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)

2. Relevant 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)

3. 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.

2. Jul 25, 2007

### PhanthomJay

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. Jul 26, 2007

### marlon

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

initial position : $$x_i = 0 e_x + 64 e_y$$
initial velocity : $$v_i = 19.8 e_x + 0 e_y$$

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

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

$$a_i = 0 t^2/2 + 19.8 t$$
$$a_y = -9.81 t^2/2 + 0 t + 64$$

Do you understand these equations ?

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

marlon

4. Jul 26, 2007

### davidatwayne

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. Jul 26, 2007

### Staff: Mentor

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:
$$y_f = y_i + v_{i,y} t - (1/2) (9.8) t^2$$

6. Jul 26, 2007

### davidatwayne

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 : )