09-Sept-2014 Trajectories

In this lab we will be using our understanding of projectile motion to predict the impact point of a ball on an inclined plane.

So for part 1, easy as cake, we figure out time using the equation with initial velocity equal to zero and acceleration equal to earth's gravity:

∆y = vt + (1/2)gt^2 = (1/2)gt^2

t = √(2∆y/g)

Using time, t, and the horizontal displacement, ∆x, we can calculate the ball's initial horizontal velocity:

∆x = vt

Now for the second part we modify the equation because the ball is now landing above ground level onto an inclined plane. 

∆x = dcos∂ = vt

t = (dcos∂)/(v)

∆y = dsin∂ = (1/2)gt^2

d = (gt^2)/(2dsin∂)

d = [(v^2)2sin∂]/[(g)cos^2(∂)]

As seen below, my calculated/predicted landing point on the inclined plane was .4647 m. The actual point of landing was .47 m. I was off by -1.12%. This error could have been from the meter stick or from the angle finder, both of which rely on my vision.