Projectile motion off a raised surface

[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

Homework 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

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.

 

[runningninja]

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

 

[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.

 

[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.

 

[Nickie]

I would like to provide a more detailed and thorough solution to this problem. Firstly, when dealing with projectile motion, it is important to understand that the motion can be broken down into two components - horizontal and vertical. The horizontal component is not affected by gravity and remains constant throughout the motion, while the vertical component is affected by gravity and follows a parabolic path.

In this scenario, since the initial velocity and angle of projection are the same, the horizontal component of velocity will also remain constant. This means that the projectile will travel the same distance in the horizontal direction regardless of whether it is shot straight or at an angle.

However, the vertical component of velocity will differ in the two cases. When shot straight, the vertical component of velocity is zero, while when shot at an angle, it is non-zero. This means that the projectile will reach a greater height when shot at an angle compared to when shot straight.

Now, coming to the question of whether the acceleration in the horizontal direction changes when shooting at an angle, the answer is no. As mentioned earlier, the horizontal component of velocity remains constant, which means that the acceleration in the horizontal direction is also zero. Therefore, the acceleration in the horizontal direction does not change when shooting at an angle.

To calculate the necessary data for the given scenario, you can use the equations of motion for projectile motion, which are:

x(t) = x₀ + v₀cosθ*t
y(t) = y₀ + v₀sinθ*t - 0.5gt²
v(t) = √(v₀² + 2g(y - y₀))

Where:
x(t) and y(t) are the position coordinates at time t
x₀ and y₀ are the initial position coordinates
v₀ is the initial velocity
θ is the angle of projection
g is the acceleration due to gravity (9.8 m/s²)

Using these equations, you can calculate the position, velocity, and time at any given point during the projectile motion. This will help you determine the necessary data to ensure that the projectile hits within 10 cm of the line drawn in the sand.

In conclusion, the acceleration in the horizontal direction does not change when shooting at an angle, but the projectile will reach a greater height when shot at an angle compared to when shot straight. Therefore, it is important to consider the angle of projection when calculating the necessary data for a projectile motion off a raised surface