Solving Parabolic Problem from Giancoli Physics 6th Edition Ch. 3

  • Thread starter doogerjr
  • Start date
In summary, the homework statement is saying that if someone shoots an arrow at a 35 m/s horizontal speed, it will take .77s to travel the distance of 27 m. The arrow will fall 3.6 meters in that time.
  • #1
doogerjr
15
0

Homework Statement


Hey all, I'm trying to get caught up in my AP Physics class, (the book we're using is Giancoli Physics 6th Edition), so I am making notes and doing the problems at the end of the chapters. (I'm on Ch. 3 right now). I understand the concepts, but can't seem to do the problems! Here is one I am stuck on;
#53 "William Tell must split the apple atop his son's head from a distance of 27m. When William aims directly at the apple, the arrow is horizontal. At what angle must he aim it to hit the apple if the arrow travels at a speed of 35 m/s?"


Homework Equations


Vy=Vyo-gt
X=Xo+VxoT


The Attempt at a Solution


Well, I drew out a diagram, and drew X and Y components for the Initial velocity (35m/s), but beyond that, I'm lost. Help appreciated!:biggrin:
 
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  • #2
You will need to break the velocity into its x,y components in terms of θ.

The time to get there will be the Vx/Distance to apple.

The time in the air going up and down is the same time, so figure from the Vy how long that would be.
(Hint: won't it be twice the time to max height where Vy is momentarily 0?)

Since the time must be equal ... then set the two equations equal.

Now you have everything but θ
 
  • #3
Apologies for stepping on your toes lowly, but i'll post anyway seen as how i spent the time to write this out!

I'd approach this problem as follows:

1. Consider the vertical component of flight. The speed at which it initially travels vertically (at t=0) will be equal to [tex]\nu=35sin\theta[/tex]. Now we know that vertically, the arrow will travel up and down in a parabolic fashion however, the total vertical distance traveled by the arrow when it hits the apple will be 0 as we are told the arrow is on the same level as the apple.

Now use the first of our equations of motion for uniform acceleration and algebraically rearrange to find the time of flight of the arrow in terms of [tex]\theta[/tex]:

[tex]& s & = ut + \tfrac12 at^2 \qquad[/tex]

where:

s represents the total distance travelled.
u represents the vertical component of velocity, in this case [tex]u=35sin\theta[/tex].
t represents the time of flight.
a represents the acceleration of the arrow (think gravity..)

2. Consider the horizontal component of flight. The speed at which it initially travels horizontally (at t=0) will be equal to [tex]\nu=35cos\theta[/tex]. Now remember, horizontally there is no acceleration! (we disregard air resistance for simplicity)

Using the equation as previous, we can simplify to find that:

[tex]& s & = ut [/tex]

Now u represents the horizontal component of velocity, in this case [tex]u=35sin\theta[/tex]. We also know s(the horizontal distance traveled given in the question) and t in terms of theta from step 1.

Put it all together, solve for [tex]\theta[/tex] et voila!
 
  • #4
Ugh. Sorry, but I don't understand.:frown:

I pretty much understand all this:
1. Consider the vertical component of flight. The speed at which it initially travels vertically (at t=0) will be equal to LaTeX Code: \\nu=35sin\\theta . Now we know that vertically, the arrow will travel up and down in a parabolic fashion however, the total vertical distance traveled by the arrow when it hits the apple will be 0 as we are told the arrow is on the same level as the apple.

From then on, I just get more steadily lost.

Edit: This really freakin annoys me cause we went over this in class months ago, and I still can't seem to remember fundamental equations or solve (relatively) simple problems.:frown:
 
Last edited:
  • #5
The simpler approach is to think it through from the basics.

You know he must launch at an angle or his son celebrates no more birthdays. That means that the velocity of the arrow to get there will be the Vx component of its velocity (35*cosθ) divided by the distance.

You can figure that time of flight though because you know Vy = g*t, is the time to get to max height, Vy = 0. Now double that for the time to return to the same height. So that means t = 2*Vy/g.

Now since the time of the 2 equations, (Vx,Vy), must be equal, the only unknown you have is the angle θ, in the trig functions that define Vx,Vy.
 
  • #6
LowlyPion said:
The simpler approach is to think it through from the basics.

You know he must launch at an angle or his son celebrates no more birthdays. That means that the velocity of the arrow to get there will be the Vx component of its velocity (35*cosθ) divided by the distance.
See, I didn't know that. Why is this?
 
  • #7
The problem tells you that the arrow is at the same level as the apple.

If the arrow shoots at 35m/s horizontally, then to travel 27 m it will take .77s.

But it will fall x in that time by

x = 1/2*g*t2 = 1/2*9.8*.6 = 2.9 m

OK. So the boy is safe, but the apple is untouched.

So you know he will only launch it at 35m/s and it will then be at an angle up to account for the time drop. So his Vx will be given by 35Cosθ and the Vy will then be 35Sinθ.

So go from there.
 
  • #8
Newton's second law

Hi doogerjr! :smile:
doogerjr said:
LowlyPion said:
the velocity of the arrow to get there will be the Vx component of its velocity (35*cosθ) divided by the distance.
See, I didn't know that. Why is this?

'cos good ol' Newton's second law :approve: says since there is no horizontal component of acceleration, the horizontal component of velocity must be constant …

so Vx is always V0cosθ :wink:
 

Related to Solving Parabolic Problem from Giancoli Physics 6th Edition Ch. 3

1. What is a parabolic problem in physics?

A parabolic problem in physics refers to a type of problem that involves the motion of objects or particles in a parabolic path, which is a curved path that resembles the shape of a parabola. This type of problem commonly involves the concepts of projectile motion or motion in a gravitational field.

2. How do you solve a parabolic problem in physics?

To solve a parabolic problem in physics, you need to identify the known and unknown variables, such as initial velocity, angle of launch, and displacement. Then, you can use equations of motion, such as the kinematic equations, to calculate the unknown variables. It is important to carefully consider the direction and magnitude of each variable to accurately solve the problem.

3. What is the difference between a parabolic problem and a linear problem?

The main difference between a parabolic problem and a linear problem is that a parabolic problem involves curved motion while a linear problem involves motion in a straight line. In a parabolic problem, the acceleration is not constant and the path is curved, while in a linear problem, the acceleration is constant and the path is straight.

4. What are some real-world applications of parabolic problems?

Parabolic problems have many real-world applications, such as calculating the trajectory of a projectile, such as a ball thrown or kicked in sports, or the path of a rocket or satellite launched into space. They are also used in designing roller coasters and other amusement park rides, as well as in analyzing the motion of objects in free fall.

5. What are some common mistakes when solving a parabolic problem?

Some common mistakes when solving a parabolic problem include using the wrong equations, not considering the direction of motion, and not properly labeling the variables. It is important to carefully read the problem and identify all the given information before attempting to solve it. Additionally, double-checking the calculations and units can help avoid errors in the final answer.

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