# Modeling Ballistics

1. Sep 6, 2008

1. The problem statement, all variables and given/known data
A child running along level ground at the top of a 30-ft-high vertical cliff at a speed of 15 ft/s, throws a rock over the cliff into the sea below. Suppose the child's arm is 3ft above the ground and her arm speed is 25ft/s. If the rock is released 10ft from the edge of the cliff at an angle of 30 degrees, how long does is take for the rock to hit the water? How far from the base of the cliff does it hit.

2. Relevant equations
V=initial velovity
t=time
s=initial height
g=gravity=32ft/s

x(t) = (V*cos(a))t
y(t) = -(1/2)g(t^2)+(V*sin(a))t+s

3. The attempt at a solution
I have identified these variables:
Variables I know:
V=25ft/s
s=3ft
a=pi/6

I also have solved 1 of the questions, time of flight by solving for this equation:
0 = -16t^2+(25/2)t+33
t=1.8789s

Now trying to use t for the range I get
x = (25* cos(pi/6))1.8789 =(approx) 40.680244

To get the distance from the base of the cliff, I subtract the distance from the edge from the range:

Distance to bas of cliff = 40.680244 - 10
But this is incorrect can anyone please help, I have been at this for hours and it seems so basic.

2. Sep 7, 2008

### Redbelly98

Staff Emeritus
That's the velocity of her arm, but you need the velocity of the rock. You'll need to combine the arm and child velocities -- and remember, they are vectors.

3. Sep 7, 2008

So do you mean that the initial velocity(V) for both equations should be 40ft/s?

4. Sep 7, 2008

### HallsofIvy

Staff Emeritus
No. The childs speed, 15 ft/sec, is horizontal only. As a vector it is <15, 0>. The initial velocity of the rock, relative to the child, is your <25 cos(30), 25 sin(30)>. Add those to find the velocity of the rock, relative to the ground.

And let's hope that child stops running before she gets to the edge of the cliff!

5. Sep 7, 2008

Right, so I get a velocity vector back how do I change this to a scalar quantity? Take the magnitude I would presume?

You know that the strange part is I got the correct answer for "Time of flight" using the initial velocity of 25ft/s.

6. Sep 7, 2008

### Redbelly98

Staff Emeritus
Treat the horizontal and vertical directions independently.

Horizontal: vi = [15 + 25 cos(30)] ft/s
Vertical: vi = 25 sin(30) ft/s

When I read the first part of the question,
"A child running along level ground at the top of a 30-ft-high vertical cliff at a speed of 15 ft/s, ..."
I thought we would be ask to calculate, coldly and objectively, her trajectory as she fell screaming onto the rocks below.

7. Sep 7, 2008

### HallsofIvy

Staff Emeritus
Which will hit first, the child or the rock?

8. Sep 8, 2008

But how do I get the rock's initial velocity if I treat the 2 vectors separate? I need a scalar quantity(not a vector) for the rocks initial velocity with respect to the earth.

9. Sep 8, 2008

O gosh I am an idiot "V" in my equation above is not initial velocity rather initial speed.
so my equations are as follows
$$x \left( t \right) = \left( \nu_0 cos \alpha \right) t \newline\newline x(t)\, is\, horizontal\, component\, of\, \mathbf{R} \left( t \right) =$$

10. Sep 8, 2008

O gosh I am an idiot "V" in my equation above is not initial velocity rather initial speed.
so my equations are as follows
$$\displaymath x \left( t \right) = \left( \nu_0 cos \alpha \right) t\, and \,y(t) = \frac{-1}{2}gt^2 + (\nu_0 sin \alpha)t +s_0$$
where x(t) and y(t) denote the horizontal and vertical components of
$$\displaymath \mathbf{R} \left( t \right) = \left[ \left( \nu_0 cos \alpha \right) t \right] \mathbf{i} + [(\nu_0 sin \alpha)t - 1/2gt^2 +s_0] \mathbf{j}$$
g is gravity = 32$$\displaymath \frac{ft}{s^2}$$
$$\displaymath s_0$$ is initial height
t is time
$$\displaymath \nu_0$$ is initial speed
$$\displaymath \alpha$$ is angle of elavation

11. Sep 8, 2008

### Redbelly98

Staff Emeritus
Looks right to me.

12. Sep 8, 2008