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**topsquark** Step 1. Draw a sketch. I can't do this, but I can explain what I have in front of me. I've got the catapult on the ground facing west so that the acceleration due to gravity will help me fire the rock the furthest. The catapult is situated on the ground and the ground is level to the point where the rock hits.

Step 2. Set up a coordinate system. There is a slightly bizzare feature here: the simplest coordinate system is not the most natural one. I will go with natural and choose a +x direction to the West and a +y direction straight up. I am going to choose an origin located at the point where the rock is fired. (Technical point here: I am assuming the rock is fired at the origin so technically the base of the catapult is underground. This is a standard assumption to make the problem simpler.)

Step 3. List all the given information. I usually use some kind of table, like the following:

t0 = 0 s................................t = ?

x0 = 0 m....x = ?..................y0 = 0 m.......y = 0 m

v0x = ?......vx = ?.................v0y = ?.........vy =?

.....ax = 3.5 m/s^2...................ay = -8.75 m/s^2

Let me explain this. We are firing the rock from the origin (the "0" variables) so x0 = 0 m and y0 = 0 m, and it is landing on the ground (which is at y = 0 m). We are firing at time 0 t0 = 0 s. We don't know when the rock lands so t is unknown. We don't know what angle we are launching at, but we can say that

v0x = (70)cos(@) m/s

v0y = (70)sin(@) m/s

where @ is the angle of inclination of the launch.

Step 4. Set up the equations.

We have 8 equations at our disposal:

x = x0 + v0x*t + (1/2)ax*t^2

x = x0 + (1/2)(v0x + vx)t

vx = v0x + ax*t

vx^2 = v0x^2 + 2ax*(x - x0)

y = y0 + v0y*t + (1/2)ay*t^2

y= y0 + (1/2)(v0y + vy)t

vy = v0y + ay*t

vy^2 = v0y^2 + 2ay*(y - y0)

We wish to maximize the range, so a likely equation to use is the first x equation:

x = x0 + v0x*t + (1/2)ax*t^2

x = 70t*cos(@) + 1.75t^2

We need to maximize x with respect to @, but we need an equation for t first. So let's look for a y equation.

y = y0 + v0y*t + (1/2)ay*t^2

0 = 0 + 70t*sin(@) - 4.375t^2

or

4.375t = 70*sin(@)

t = (70/4.375)*sin(@)

Putting this into our x equation:

x = 70t*cos(@) + 1.75t^2

x = 70[(70/4.375)*sin(@)]*cos(@) + 1.75[(70/4.375)*sin(@)]^2

x = 1120*sin(@)*cos(@) + 448*sin^2(@)

To find the max x we need to vary this with respect to @. So take the first derivative of x with respect to @ and set it to 0:

0 = 1120*cos^2(@) - 1120*sin^2(@) + 2*448*sin(@)*cos(@)

Analytically speaking, this is complicated to solve. If you are allowed, I would recommend you use a calculator to solve this. Otherwise I would start by noting that @ is in the first quadrant, so we may put:

cos(@) = +sqrt(1 - sin^2(@))

So the equation is:

0 = 1120*(1 - sin^2(@)) - 1120*sin^2(@) + 896*sin(@)*sqrt(1 - sin^2(@))

2240*sin^2(@) - 1120 = 896*sin(@)*sqrt(1 - sin^2(@)) <-- Square both sides.

(2240*sin^2(@) - 1120)^2 = 802816*sin^2(@)*(1 - sin^2(@))

5017600*sin^4(@) - 5017600*sin^2(@) + 1254400 = 802816*sin^2(@) - 802816*sin^4(@)

5820416*sin^4(@) - 5820416*sin^2(@) + 1254400 = 0

To make this look a little simpler, set z = sin^2(@):

5820416*z^2 - 5820416*z + 1254400 = 0

z^2 - z + 0.215517 = 0

Using the quadratic equation:

z = (1 (+/-) sqrt((-1)^2 - 4*1*0.215517))/(2*1)

z = 0.685695 or z = 0.314305

So

sin(@) = (+/-)sqrt(z)

sin(@) = 0.82867 or sin(@) = 0.560629 <-- I'm only choosing the + solutions since @ is in QI.

Thus @ = 55.9007 degrees or @ = 34.0993 degrees.

As to what the plot looks like, look at the x equation again:

x = 70t*cos(@) + 1.75t^2

@ is just a constant, so this will be a parabola on a t vs. x plot.

-Dan