How Do I Correctly Calculate Height Over Time for an Object on a Ramp?

[mastermind1]

Homework Statement

time = 1 seconds, 2 seconds, 3 seconds, 4 seconds, 5 seconds
acceleration due to gravity = -9.8 m/s^2
Max. Height = 3 meters

Homework Equations

d = vit + 1/2at^2
v = d/t

The Attempt at a Solution

I actually have to create values myself, so that the as the time increases the height decreases.
This is a scenario where a golf ball on a ramp from the top of the ramp is released, and for each of those time values I mentioned above I need the height.

My attempt:

v = 3m/1s = 3 m/s
d = (3m/s)(1s) + (1/2)(-9.8m/s^2)(1s)^2 = -1.9 m

If I keep increasing the time values my height is going to increase. That is not what i want. Please help.

Thanks.

 

[eHoaX]

Your max. height is too low. In order for this too work it needs to be much greater than 3 meters.

 

[mastermind1]

I just chose 3 as a random number. It can be anything. Help me. Choose whatever number you want but try to keep it as low as possible.

And, why is it not possible with three. Practically, that is what is happening. As the time increases the height decreases. The ball is initially at rest before it is released from the top of the ramp.

 

[eHoaX]

OK assuming the ball goes into a freefall after it is released from rest.


It is not possible with 3 meters because the ball would hit the floor before 1 second time, it would travel more than 3 meters in free fall.

Lets try 130 meters above ground level on a super high ramp.

So your intitial values are as follows:

Δy=height(after "x" seconds)-initial height
initial velocity = 0 (at time = 0)
time = 1,2,3,4,5
acceleration = -9.8

now substitute this into the kinematic equation and the only thing that you should change when solving for height(after "x" seconds) should be the time.

 

[Student100]

OK assuming the ball goes into a freefall after it is released from rest.


It is not possible with 3 meters because the ball would hit the floor before 1 second time, it would travel more than 3 meters in free fall.

Lets try 130 meters above ground level on a super high ramp.

So your intitial values are as follows:

Δy=height(after "x" seconds)-initial height
initial velocity = 0 (at time = 0)
time = 1,2,3,4,5
acceleration = -9.8

now substitute this into the kinematic equation and the only thing that you should change when solving for height(after "x" seconds) should be the time.

Maybe I'm a bit confused by the way the question was asked, but I'll say a couple of things here:

1. I hate when problems or teachers force you to use their coordinate system. The problem should state simply g not -9.8.

2. An object on a ramp isn't in free fall. Gallieo used a ramp to time objects more accurately then he could by dropping them. Think about it this way, use the fundamental theorm of calculus on an object with constant acceleration. Now take both intial velocity and initial position to be zero. Since you're on an inclined plane, the force of g will be split into two perpendicular components. With one component parallel to your ramp. So the acceleration isn't g, but some new acceleration derived from g. 3 meters is plenty long enough. Look up gallieos experiment if you're still confused.

 

[NascentOxygen]

If you use a ramp of 3m having almost no slope then the ball could take a couple of minutes to meander its way to the lower end!

Draw a body on an inclined slope, and show the triangle of forces that allows you to resolve its weight into two components, one being parallel to the slope, the other being perpendicular to the slope. Only the component parallel to the slope can cause any change in the body's motion.

F = ma then tells you the acceleration along the slope, and you can apply your kinematics equations to its motion.