Topic: HW 3a #2
A ball is thrown off a 36 m high cliff at 3 m/s, 50o above the horizontal. How long does it
take to hit the ground?
Student A: for question # 2 on HW 3a, i'm not sure i will be able to solve for this
equation because i have so little information. to solve for time, i would need at least the
displacement of the cannonball.
Dr. Man: First, knowing which direction of motion you are working on.
Which direction of motion limits the time? Horizontal direction or vertical direction?
In the Vertical direction, what was the displacement Yf-Yi ?
(choose positive Y direction first).
Do you know v0y, the initial velocity in the y direction? You know the total v0 and you
know the angle, you should be able to find its y direction component.
Once you know initial v0y, delta y, ay in the y direction, it is easy to find time before
landing using your knowledge from chapter 2.
Student B: I have chosen the positive direction to be upward. I have computed the
Voy to be 2.298m/s in the y direction. delta_X = Xf-Xi ...delta_X=-36m
Vy= ?
t=?
a=-9.8m/s^2
I am using the equation:
delta_X = Voy(t) + 1/2at^2
-36 = 2.298t + 1/2(-9.8)t^2
-2.298t-36 = -4.9t^2 (divide everything by t)
-2.298-36 = -4.9t
-38.298 = -4.9t
t= 7.816s -WRONG
I feel like all my variables are correct and I've checked the math. What am I doing wrong?
Dr. Man: Good start. But first, just don't mix X and Y directions.
You meant that Delta_Y= Yf - Yi= - 36 m
delta_Y = Voy(t) + 1/2at^2
-36 = 2.298t + 1/2(-9.8)t^2
-2.298t-36 = -4.9t^2 (divide everything by t)
Here: if you divide everything by t, what will you do with 36/t?
For this equation, you need to arrange it to look like:
at^2+b^t+c=0
a, b and c are numbers and can be positive or negative.
Then you use quadratic equation formula to solve t.
Again, now we all see how important the math skills we tested in the readiness exam was
Student A: Thank you for your response. I actually tried using the quadratic formula first,
and put in both the +/- versions of time. I tried it again now, using the same numbers and
got an answer that was within 1% of my original answer, which was wrong according to
webassign. Hmm, I'll have to try it again when I have more time to digest the information.
Dr. Man: Make sure that you get the RIGHT numbers for a, b and c in your equation!
A+B=C
when you move B to the other side, positve become negative: A=C-B
A-B=-D
When you move the negaticve D to left side it becomes positive.
so: A-B+D=0
at your equation a*t^2+b^t+c=0 correctly first!
Student B: so for this problem, we would use the quadratic formula instead of using t=2
(initial velocity of y)/g?
Dr. Man: Never memorize conclusions from other questions or special cases.
Whatever equation to use, you should have a clear idea why that equation is valid for your case.
For a canon ball shooting from ground level and falling to the ground level (same height
as initial position, using t=2(initial velocity of y)/g is correct.
But DO YOU UNDERSTAND WHY? Do you know why there is a factor of 2?
For cannon ball falling to the same vertical height, Delta_Y= Yf -Yi =0
As a result, in the y direction Vfy^2 = Viy^2 + 0
The initial and final y direction SPEED is the same.
We know Vfy flipped direction comparing with Viy.
So Vfy=-Viy Vfy - Viy = -2Viy
t= (Vfy-Viy)/a = -2Viy/ (-g) = 2Viy/g
Got it? The reason for t = 2Viy/g, this case is that it fall to the same height, and Delta_y=0.
hence Vfy^2=Viy^2
On the other hand, The time to reach maximum height t_top should be found in the equation that
V-top = Viy -g* t_top.
So, t_top = (Viy - V_top)/ g =Viy/g
If you fall to the same height, the falling part is SYMMETRIC to the raising part, as a result
the total time is 2*Viy/g
Again, it is absolutely wrong to memorize equations and apply them to other problems.
For example, in 3a #3, it fall to another height. Delta_y is not zero. Raising and falling
part are definitely not symmetric. t=2*Viy/g is wrong.
Actually, HW 2b, number 4, the Diver question is the same kind of y direction as ad 3a #2.
To study that HW solution first, and to make sure you have no problem with that
before last weekend, will make it so much easier for you to finish this HW.
Topic: HW3a #3 This question will apprea again in HW 3b. Answered will be disclosed
in HW 3b solutions.
An astronaut on the planet Zircon tosses a rock horizontally with a speed of 7.75 m/s.
The rock falls through a vertical distance of 1.40 m and lands a horizontal distance of 8.95 m
from the astronaut. What is the acceleration of gravity on Zircon? (Neglect air resistance.)
Student A: I'm not sure how to solve this, any help?
Dr. Man: for this question you also need to understand lecture sample of ball B tossed horizontally first.
The difference is that the acceleration in that planet is not known. you need to solve "a" this time.
you should know delta_y, Vy initial, and you should be able to find t from what are told in x direction.
(you know the initial speed in x direction and the horizontal flying distance)
Student B: I am still not sure how exactly you could find the time if you dont know the acceleration.
Im trying to find time by using the equation: t^2=2(1.40)/-9.81 but we cannot use this
right because we dont know a? im not even sure if im setting this up right for time.
Dr. Man: Read the detailed explanation on the topic of 3a #3 FIRST.
Make sure you understand all solutions in HW 3a before continue.
Again, as I said in answering about 3a #3.
t=2 V0y /a is not applicable for many cases.
And what you wrote:
t^2 =2(V0)/a is never true.
1, TREAT x and y direction completely separately.
Decompose V0 into x and y directions.
V0y =
V0x =
Initially was sent out horizontally, V0y =0
V0x=V0
2,For the whole flight , what do you know for the y direction motion?
a_y? (it's not on earth)
V0y (initial velocity in y direction is known)
Vfy? (it hits the ground with a speed)
Delta_Y (vertical displacement was known)
t?
Ok, we don't know 3 of them. It's hard to solve.
3, What are other information we have?
For the whole flight. What do you know about x direction motion.
Delta_X (we know horizontal displacement)
V0x (we know initial speed in x direction)
Are there any acceleration in horizontal direction?
(any force in horizontal direction during the flight?)
So, can you find time t from Delta_x and V0x?
Once you have time t, can you go back to step 2 and find out a_y?
Which equation you need in order to find a_y, if you know V0_y, Delta_y, and t.
Start from the very beginning. Step by step.
Understand chapter 2 before doing problems in chapter 4.
Topic: HW3a question #4
A diver runs horizontally off the end of a 2 m high diving board at 4 m/s.
a.) How long does it take for the diver to hit the water?
b.) How far horizontally does the diver end up from the board?
c.) What is the size of the horizontal (x) component of the diver’s velocity just before he hits the water?
d.) What is the size of the vertical (y) component of the diver’s velocity just before he hits the water?
e.) What is the speed of the diver just before he hits the water?
f.) If the diver instead horizontally ran off the end at 12 m/s, which of the answers for a-e would change?
Student A: I found the time but I don't know which formulas to use to find the other answers.
Please help
Dr. Man: It helps to first read the lecture notes one more time and read the sample questions
on the book before doing this problem. Key points: Treat horizontal and vertical
direction motion separately. X direction and Y direction variables do not affect each
other. The only thing x direction and y direction share is time t. In y direction, constant
acceleration due to gravity. In x direction, no acceleration. Object simply keeps its
initial x-component velocity. Constant velocity in x direction! Vx doesn't change!!!
Which direction of motion limits the travel time? (you already figured it out and finished
part a) Next, Which direction of motion decides the distance of travel in the horizontal
direction? How far did it move in the x direction? (you already found the time!)
c, asks you Vx at the end.
d, asks you Vy at the end. Which direction of equations to use?
What are really known and which equation can solve Vy_final for you?
e, asks you the total magnitude of V.
once you have Vx and Vy, how to find the Size of the vector V? (all on lecture note and on the book).
Again, see HW first and read lecture notes and textbook (sample questions and chapter summaries)
carefully with questions in mind. Then you can do HW without problems.
Topic: HW3b #2
In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for
accuracy in hitting a target on the ground (Figure 4-18). Suppose that the tower height is
h = 9.20 m, and that the bulls-eye's horizontal distance is d = 3.6 m from the launch point.
If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye?
Student A: How do we solve this?
Dr. Man: this is exactly like the sample question on lecture note.
(Ball B, being thrown horizontally).
In the sample question we solved during class,
1, we know the height, we can find the falling time.
2, we know the the initial velocity in horizontal direction, we can find the distance it moved
horizontally during this motion.
The first part is the same. then this question just told you the horizontal distance and ask
you to find the needed velocity in horizontal direction
Topic: HW3b#4 to #8 are reading assignment related to chapter 5.
Topic: HW3b#8
Student A: I thought an action-reaction pair was produced by physical contact of two
objects (the brick and my hand) ......Action-reaction pair or not? Why is this?
Dr. Man: Please read the part about Newton's third law more carefully.
Action-reaction pair are about TWO objects: A does something to B and B must do
something to A. You PUSH table, The table RESISTS you. These two forces are
Action-reaction pair. The earth attracts you with gravity and you also attract the earth.
They are Action-reaction pair. Key points: Two objects, Switching roles in these actionreaction pairs.