If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t) = 32 + 80 t - 16 t^2. What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?
I don't want the answers but more on the method of how I would be begin to solve this? Like what steps I should do?
Posted by Bokonon (Member # 480) on :
Well, the [EDIT: first] solution would be a maxima, as the equation is graphed, right? In other words, [EDIT: graphing] height on the y-axis and time on the x-axis, the solution would be the highest value of y plotted on the graph.
Integrals, if you recall, can be used to determine the area under the curve.
Derivatives can be used to determine the slope of a curve at a given point.
That's a very generic start.
-Bok
Posted by swbarnes2 (Member # 10225) on :
The velocity is the change in position.
You know how to use calculus to derive the equation for the change in position as a function of t. In fact, if you just started calculus, it's probably the only calculs trick you know.
And remember at the ball's max height, its velocity is 0.
Posted by Mucus (Member # 9735) on :
Its a trick question. The velocity of the ball is 0, hitting the ground tends to decelerate things.
Posted by dantesparadigm (Member # 8756) on :
quote:Originally posted by Mucus: Its a trick question. The velocity of the ball is 0, hitting the ground tends to decelerate things.
Not quite. It's asking for the velocity just before the ball hits.
Posted by Shigosei (Member # 3831) on :
If it was asking for the velocity just before the ball hits then it would say that. Since it doesn't actually say that, its clearly asking for something else.
Posted by Blayne Bradley (Member # 8565) on :
I can do differentiation, but we havent done integration or maxima extreme yet or any graphing.
Posted by Blayne Bradley (Member # 8565) on :
"0" for part b would appear to be incorrect.
Posted by scifibum (Member # 7625) on :
Mucus, you could be right, but you're probably not. It's EXTREMELY common for questions like that to be written ambiguously. When I prep for IT certifications, it drives me insane: I miss questions because they were worded ambiguously. I can't get a practice score that equates to a passing score on the real test. Then I go take the real test, where they actually try to make most of the questions have one and only one right answer, and I pass fine.
Posted by Icarus (Member # 3162) on :
No, the velocity is not zero. Calculus is all about the instantaneous (as opposed to the averaged). At the *instant* the ball hits the ground, it is moving with a velocity. The ground puts a stop to it, but the question is about that very instant. It's not a trick question--it's very much calculus bread and butter.
Posted by Mucus (Member # 9735) on :
Best to verify your assumptions and email the prof.
Ask him whether they want the "velocity just before the ball hits" or when it actually does hit. You can probably work a limit into there.
Don't forget to also ask about the radius of the ball. You probably can't assume that the ball is a massless point, otherwise it would be pretty hard to throw. If the equation describes the middle of the ball, then you'll need to account for the radius when it touches the ground.
Posted by Icarus (Member # 3162) on :
um, trust me on this. You may assume the ball is a massless point. And they want the rate of change at the moment the ball's height is zero. This is a standard problem.
Posted by Mucus (Member # 9735) on :
You try gripping and throwing a massless point.
Posted by Icarus (Member # 3162) on :
I suppose you'll also encourage Blayne to e-mail his prof and find out if the ball is being thrown in a vacuum . . .
Posted by Mucus (Member # 9735) on :
Well thats just silly.
The equation describes the velocity of the ball, even if it was in a vacuum the lack of air resistance should be accounted for already in the equation.
Posted by Blayne Bradley (Member # 8565) on :
I have to agree with Icarus, trick questions have never been in my math with cal. They want the volocity just as it hits (which still possesses velocity otherwise it wouldn't bounce).
I can derive, like 32+80t-16t^2
becomes 80-32t
second derivitive is -32
But I have no idea what I am supposed to plug in or rearrange at all to get me whats the maximum height the ball travels.
Posted by Icarus (Member # 3162) on :
When the ball is at its maximum height, its velocity is zero.
And the velocity is given by the derivative.
Posted by Blayne Bradley (Member # 8565) on :
32+80(2.5)-16(2.5)^2 woot for part a.
Part B is troubling me now.
Posted by Icarus (Member # 3162) on :
Can you find at what time the ball hits the ground?
Posted by Blayne Bradley (Member # 8565) on :
ya I had to take the original function 32+80t-16t^2 factor it to -16(t^2 - 5t -2 ) then use a quadratic formula of -b -/+ ( b^2 + 4ac ) ^ 1/2 / 2a to get my time which I put back into the derivitive of f(x).
Posted by Icarus (Member # 3162) on :
(B squared minus four AC)
Posted by Blayne Bradley (Member # 8565) on :
typo. Basically we have webwork, which is an online way of subitting assignments 1 question at a time.
Posted by swbarnes2 (Member # 10225) on :
quote:I can derive, like 32+80t-16t^2
becomes 80-32t
But you have to understand what you did, and what it means.
The derivitive equation is the one that describes the change of the previous equation as a function of time.
And what is the significance of knowing the exact rate of change in position? The rate of change in position is the velocity.
quote:second derivitive is -32
Right...this is the rate of change of the velocity. And what does that mean? It's the acceleration. 32 ft/sec/sec. Sound familiar?
If you know physics, you don't even have to use the quadratic equation. Potential energy = mgh. kinetic enegy = .5mv^2. So pick some point at which you know both the height and the velocity of the ball, work out the total energy, and at p = 0, all that energy is kinetic.
[ October 18, 2008, 05:21 PM: Message edited by: swbarnes2 ]
Posted by Nighthawk (Member # 4176) on :
What if it's a bouncy ball? Then the velocity wouldn't be zero at height zero; it would be reflected upwards off the ground.
So in order to answer the question, we need to know the elasticity of the ball to determine the loss in velocity due to impact.
Posted by BlueWizard (Member # 9389) on :
I can't analyze it from a calculus perspective but from a logical perspective, as the ball is thrown upward with an INITIAL velocity of 80ft/sec, it will begin to decelerate relative to gravity. When something falls or rises in gravity, it doesn't move or change at a constant rate; it accelerates or decelerates, which is a nonlinear rate of change.
So, apply gravitational deceleration to the ball until the velocity is zero. From that, you should be able to determine it's height at that point.
From that peak of apogee, the ball with then start to fall from this now known distance, and accelerate at the speed of gravity. Knowing the height and the rate of change of the speed, you should be able to know the time to the ground and the speed when it strikes the ground.
However, are we going to consider Terminal Velocity? Gravitational acceleration, is under ideal circumstance; in other words, in a vacuum. Because Air is a dense fluid, it exert a counter force to gravity, meaning that the ball will accelerate until its downward force is equal to the resistance of air, at which time its speed will be constant.
So, when the ball move upward, we have a negative rate of change over time; meaning deceleration. When deceleration reaches zero, the ball is stopped at a predictable height.
Then that ball starts to fall. We now have a positive rate of change over time; meaning acceleration. The rate of change is constant and predictable, the height is known, and so we can predict the time to the ground, and the speed when it reaches the ground.
All Calculus is, is predicting nonlinear rates of changes over time.
Probably not much help, but there it is.
steve/bluewizard
Posted by The Rabbit (Member # 671) on :
The equation BB has is based on the presumption that there is no air resistance but you really don't need to understand that to solve the problem as it is given. Its also clear that the equation only valid for the time period between when the ball is thrown (t=0) and when it makes contact with the ground.
The maximum height of the ball will be at the point where the velocity is zero. So you take the time derivative of the equation and set it equal to zero to find the time when the velocity equals zero, then you plug that back into the original equation to get the position at that time.
Part be is similar. You solve the original equation for s = 0 to find the time when the ball hits the ground. There will be two solutions to this equation but only one of them will be physically meaningful. The other will be for a time less than zero when the equation doesn't apply. You then take this time and plug it into the velocity equation (i.e. ds/dt) to get the velocity when the ball is at s = 0.
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