Calculate velocity with initial height, range and angle

[spiderdan]

Homework Statement

My lab partner and I were to find the optimum angle a nerf gun could be fired at to provide the longest distance from the starting point. We also need to calculate the initial velocity of the projectile and the range and height the projectile should have reached for each angle. Air resistance is ignored. The data my partner and I collected in the lab was as follows

Homework Equations

v = v₀ + at

x = x₀ + v₀t + 1/2at²

v² = v₀2[/sub] + 2a(x - x₀)

note: apparently, these are the four equations which are given on the AP exam. I am allowed to use other projectile motion equations, but I must show how they were derived from these three.

The Attempt at a Solution

I was able to solve at zero degrees quite easily.
y = v_0yt + 1/2gt² + y₀ (from second equation)
y = (v_0ysinθ)t + 1/2gt² + y₀
0 = (v_0ysin0)t + 1/2*-9.8m/s²*t² + .84m
0 = -4.9m/s²*t² + .84m (sin0 = 0)
t = .414

x = x₀ + v₀t + 1/2at²
x = v_0xt + 1/2at²
x = (v₀cosθ)t
6.24m = (v₀cos0)*.414s
v₀ = 15.07m/s

When the angle becomes nonzero is where I run into trouble. I can't solve for time and then plug into the other equation for velocity. There is an equation in my physics book which seems to be for this very purpose.

R = (v²₀/g)sin2θ₀

The problem is that I can't use an equation straight out of the book, I have to derive it. That, and that it doesn't have Y₀

Using the book equation for 15°

10.17m = (v²₀/9.8m/s²)sin2(15)
v²₀ = 14.12

I had the thought that I might be able to solve one equation in terms of v₀ and then plug this into the other equation. I tried this leaving Y₀ at 0 to see if it matched up with the books equation

0 = (v₀sin15)t - 4.9t²
4.9t² = (v₀sin15)t
4.9t = v₀sin15
18.93t = v₀
t = v₀/18.93

10.17m = (v₀cos15)(v₀/18.93)
192.53 = (v₀cos15)v₀
v²₀ = 199.32
v₀ = 14.12

This works, but if I bring Y₀ into the picture, the equation becomes much more difficult to solve. Here is my attempt.

y = y₀ + V_0y + 1/2gt²
0 = .89m + (v₀sin15°)t - 4.9 m/s²t²
0 = -4.9 m/s²t² + (v₀sin15°) + .89m

I then plugged this into the quadratic formula and got t = .00140v₀² - .182

10.17m = (v₀cos15°)(.00140v_0]² - .182)
0 = .0013524v₀³ - .175812v₀ - 10.17
v₀ = 21.1795 m/s

this velovity is faster than the one with a Y₀ of 0, which can't be right. Where am I going wrong?

As a side note, I believe it was assumed the gun would be held on the floor, making the slight y₀ irrelevent. My group fired the gun from a table, which obviosly makes y₀ more important. Is there some way we can add y₀ into the equation, or do we just need to account for this in our error analysis? Thanks ahead of time!

 

[haruspex]

Although the image is lost, I deduce from your working that the target height is 0.84m (or is it 0.89?) below the launch height. That means the range equation R = (v²₀/g)sin2θ₀ is not valid anyway.

plugged this into the quadratic formula and got t = .00140v02 - .182

There should be a square root in there somewhere.
To avoid that, just write out the horizontal and vertical displacement equations as functions of t and v₀. Since you want the max horizontal displacement, $\frac{dx}{d\theta}=0$. Differentiate both displacement equations wrt $\theta$.
See if you can take it from there.