1. ## Vector Geometry: Projectiles

Hi guys, I am working on some Vector Geometry homework and have come across a problem that I don't know what to do for, any help would be greatly appreciated:

A spring catapult slings a boulder in the air from ground level with an initial speed of 280 feet per second and the boulder crashes back into the ground 7.7 seconds later. At what angle of elevation in degrees was the boulder slung into the air (in degrees)?

I know you would start with a = <0, -32> and v = <280cos(x), 280sin(x)> but I'm not quite sure where the problem goes from there, any help would be greatly appreciated.

2. ## Re: Vector Geometry: Projectiles

Originally Posted by colinvt11
Hi guys, I am working on some Vector Geometry homework and have come across a problem that I don't know what to do for, any help would be greatly appreciated:

A spring catapult slings a boulder in the air from ground level with an initial speed of 280 feet per second and the boulder crashes back into the ground 7.7 seconds later. At what angle of elevation in degrees was the boulder slung into the air (in degrees)?

I know you would start with a = <0, -32> and v = <280cos(x), 280sin(x)> but I'm not quite sure where the problem goes from there, any help would be greatly appreciated.
Call your variable $\theta$ to put it in the usual format.

There are a number of ways to approach this. The simplest is to use the equation
$\overrightarrow{s} = \overrightarrow{s_0} + \overrightarrow{v_0}t + \frac{1}{2} \overrightarrow{a} t^2$

where s is the final displacement <r, 0> (r is the range, an unknown), s_0 is the initial position <0, 0>, v_0 is the inital velocity, and a is the acceleration. This equation decouples into x and y equations. What you want is the y equation. This only has $\theta$ as a variable.

-Dan

PS: This should be in the Math subforum.

3. ## Re: Vector Geometry: Projectiles

Hello, colinvt11!

A spring catapult slings a boulder in the air from ground level with an initial speed of 280 ft/sec
and the boulder crashes back into the ground 7.7 seconds later.
At what angle of elevation was the boulder slung into the air (in degrees)?

The equations for projectile motion (for this problem) are:

. $\begin{array}{ccc}x &=& (280\cos\theta)t\qquad\;\; \\ y &=& (280\sin\theta)t - 16t^2 \end{array}$

$\text{When }\,t = 7.7,\;y \,=\,0.$

$\text{We have: }\:(280\sin\theta)(78.7) - 16(7.7^2) \:=\:0$

. . . . . . . . . . . . . $2156\sin\theta - 948.64 \:=\:0$

n . . . . . . . . . . . . . . . . . . . . . $\sin\theta \:=\:\frac{948.64}{2156} \:=\:0.44$

n . . . . . . . . . . . . . . . . . . . . . . . $\theta \:\approx\:26.1^o$

-Dan

5. ## Re: Vector Geometry: Projectiles

Thank you both very much, I was able to understand the problem from your descriptions!

6. ## Re: Vector Geometry: Projectiles

Suppose that the equation of the line is:
r = (a1, b1, c1) + s(d1, d2, d3), where s is real

Suppose that the given point is: (a2, b2, c2)

We obtain the direction vector by considering the 2 points, e.g. (a2 - a1, b2 - b1, c2 - c1) = (e1, e2, e3). (a1 - a2, b1 - b2, c1 - c2) is also alright.

Equation of the plane is thus:
r = (a1, b1, c1) + s(d1, d2, d3) + t(e1, e2, e3) where s, t are reals.

or

r = (a2, b2, c2) + s(d1, d2, d3) + t(e1, e2, e3) where s, t are reals.

If you like to obtain the equation in scalar product form, it will be: r . n = a . n where
n = (d1, d2, d3) x (e1, e2, e3) and a = (a1, b1, c1) or (a2, b2, c2).