# Calculate velocity with initial height, range and angle

## 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 = v0yt + 1/2gt2 + y0 (from second equation)
y = (v0ysinθ)t + 1/2gt2 + y0
0 = (v0ysin0)t + 1/2*-9.8m/s2*t2 + .84m
0 = -4.9m/s2*t2 + .84m (sin0 = 0)
t = .414

x = x0 + v0t + 1/2at2
x = v0xt + 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 cant 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 + V0y + 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°)(.00140v0]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!