# Projectile motion off a raised surface

1. Mar 17, 2013

### KarlBu

1. Easy solution that I'm pretty sure I talked myself out of.
Using a spring loaded shooter, first shoot it off a table and calculate data and then move to another table of a different height and shoot with an incline into a sandbox. Must hit within 10cm of line drawn in the sand.
known info - y(f) = 1.62 m from center of ball to second platform, θ = 27 °
f(g) = 9.8 m/s/s, using same shooter described in 2

2. Relevant equations
information from straight shot from a table at .86m

Initial Final
x₀= 0 x = 2.925 m
y₀ = 0 y = .86 m
v₀ = 6.98 m/s v = 6.98 m/s
v₀ = 0 v = 4.11 m/s
t₀ = 0 t = .419 s
a = 0
a = 9.8 m/s/s

3. The attempt at a solution

The only thing i need to know is if acceleration in the x direction would change (no other different factors) when changing from shooting in a straight line versus shooting up in the air at an angle. I know the different velocity equations (Vsinθ) so no problem their. I do not think it does change as using same spring but, I started to question myself when thinking about the parabolic shape of the projectile motion along the x axis when shot with a 27 degree up angle.

Last edited: Mar 17, 2013
2. Mar 17, 2013

### runningninja

Assuming a standard coordinate system, the x component of the velocity still changes at a constant rate, meaning the x accleration remains constant.

3. Mar 17, 2013

### KarlBu

still not quite grasping concept

it is a standard coordinate system. constant velocity (6.98 m/s) with 0 acceleration (in the x direction) when shooting straight. When shooting up 27 degrees would that cause a change in acceleration (from a=0 shooting straight) to appear in the x component due to angle versus x direction? If so how would i go about solving for it? (i would assume something like Vcos(θ)?? if acceleration is some other constant than 0.)
known info - y(f) = 1.62 m from center of ball (at release point) to ground, θ = 27 ° up
f(g) = (m)9.8 m/s/s, using same shooter described in 2, information from shooting straight listed above.

4. Mar 17, 2013

### KarlBu

Solved. Unless another force acting upon it (ie rocket booster) acceleration along the x component will be 0 regardless of angle. The only force acting upon it will be gravity which only affects the y component.