# Kinda embarrassing to ask about setting up this simple kinematics eqtns

• flyingpig
In summary, the problem involves a biker leaving a ramp at an angle of 30 degrees with an initial velocity of 20m/s. The goal is to set up the motion equations, find the maximum height reached, and determine when the biker lands on top of a new building. The motion equations are y(t) = -5t^2 + |20|sin30t + (10 + 20) and x(t) = |20|cos30t. The maximum height reached is 35m. The biker lands on the building when t = 1 + √5. The angle of the velocity vector is phi because at the instant the biker leaves the ramp, the angle of the velocity vector
flyingpig

## Homework Statement

I made this one up and now i can't solve it (tada...)

[PLAIN]http://img577.imageshack.us/img577/5081/probloem1.jpg

OKay so the biker runs off the ramp and leaves the ramp at the tip with some initial velocity v0

I will give some numerical values to make this easier

ϕ = 30 degrees
g = -10m/s2
v0 = 20m/s
y = 10m
h = 20m
d = 5m
x = 30m

OKay three goal

1) Set up the motion equations for the biker
2) Find the max height reached
3) Find when it lands on top of the new building

## The Attempt at a Solution

1) I have most problems with this one, especialyl when I cannot decide if ϕ is the correct angle. That is if the biker flies off from an angle different from ϕ

$$y(t) = -5t^2 + |20|sin30t + (10 + 20)$$

$$x(t) = |20|cos30t$$

2)

$$y(t) = -5t^2 + |20|sin30t + (10 + 20) = -5t^2 + 10t + 30 = -5(t^2 - 2t - 6) = -5(t^2 -2t + 1 - 1 - 6) = -5((t -1)^2 - 7) = -5(t-1)^2 + 35$$

So ymax = 35m

3) So basically h - d = 20 - 5 = 15

$$15 = -5^2 + 10t + 30$$

$$0 = -5t^2 + 10t + 20$$

$$t = 1 + \sqrt{5}$$

Last edited by a moderator:
Close, you should have a +15 in the second to last equation, not 20.

Also, instead if figuring out the whole equation for a parabola with that silly math you did, you can immediately figure out when the biker reaches the top of the parabola because you know $v_{y}(t) = v_0 - at$ and you simply set the velocity to 0 and you know immediately what time you reached the top.

Pengwuino said:
Close, you should have a +15 in the second to last equation, not 20.

Also, instead if figuring out the whole equation for a parabola with that silly math you did, you can immediately figure out when the biker reaches the top of the parabola because you know $v_{y}(t) = v_0 - at$ and you simply set the velocity to 0 and you know immediately what time you reached the top.

It's called being lazy...

Just wondering why is using phi right? I thought it should have left with a different angle than phi

At the instant that the biker leaves the ramp what is the angle of the velocity vector? that is whi phi is the right angle to use. Also, why do you think that it should be some other angle?

wbandersonjr said:
At the instant that the biker leaves the ramp what is the angle of the velocity vector? that is whi phi is the right angle to use. Also, why do you think that it should be some other angle?

What if he doesn't actually fly off? But instead trips with that velocity?

flyingpig said:
What if he doesn't actually fly off? But instead trips with that velocity?

Well that's like saying what if he gets hit by a meteor instead of flying off.

He's going to fly off at that angle because there is no other angle he could possibly fly off at due to inertia

Actually something still doesn't make sense, it came to me last night.

$$y(t) = -5t^2 + |20|sin30t + (10 + 20)$$

This says that the biker is subjected to gravity even before he leaves the ramp, but when he is on the ramp, there is a normal force, so he isn't free-falling like the equation says. This equation is only valid when he leaves the ramp no?

The equation is only valid after the person has left the ramp.

If you want to calculate the equations of motion while he's on the ramp, he wouldn't be in free fall and you'd have to take into account the force he exerts to push himself up the ramp and the normal force and all that.

But that equation is a parabolic path, how do I know at what value of t does the equation become valid?

By convenience you start the problem at t=0. Whatever happens when the guy is on the ramp is taken care of by the fact that you know the end velocity and the angle he flies off at.

I just graphed it and it was so consistent that I am in disbelief.

## 1. What are kinematics equations?

Kinematics equations are mathematical expressions that describe the motion of objects in terms of their position, velocity, and acceleration. They are used to solve problems related to motion and can be applied to a variety of physical systems.

## 2. How do I set up kinematics equations?

To set up kinematics equations, you need to identify the known and unknown variables in the problem. You also need to determine which kinematics equation(s) to use based on the given information. Once you have all the necessary information, you can plug the values into the equation(s) and solve for the unknown variable.

## 3. Can I use kinematics equations for any type of motion?

Yes, kinematics equations can be used for any type of motion as long as the acceleration is constant. This includes linear motion, projectile motion, circular motion, and rotational motion.

## 4. Do I need to memorize all the kinematics equations?

No, you do not need to memorize all the kinematics equations. It is more important to understand the concepts behind the equations and know how to apply them to solve problems. However, it is helpful to be familiar with the most commonly used equations.

## 5. What are some common mistakes when setting up kinematics equations?

Some common mistakes when setting up kinematics equations include not correctly identifying the known and unknown variables, using the wrong equation for the given problem, and not paying attention to units. It is important to double-check your work and make sure all variables are consistent before solving the equation.

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