Question

Can someone help with this?
http://oi63.tinypic.com/2v0z5ew.jpg

Answer 1

@CShrix

Answer 2

Keep in mind that the vertical and horizontal velocities are going to be the same. This is because the components of the magnitude of the velocity at 45deg is the same.
AKA \(\cos(45)=\sin(45)\)

Answer 3

.71=.71

Answer 4

@ganeshie8

Answer 5

Consider the velocity in vertical direction first

Answer 6

whats the vertical component of velocity when the ball is at its maximum height ?

Answer 7

At the top where it has the V

Answer 8

At the maximum height it has only horizontal component of velocity

btw @ganeshie8 not according to mathematical equations but according to theory how can we say that there is no vertical component of velocity at the highest point?

can we consider it says that it as at the time of reversal v=0 ?

Answer 9

i don't know, but it seems to be a general conclusion that a nice continuous function reaches a max/min when its derivative (rate of change) is zero.

Answer 10

I think it is nature's way of conserving energy. At any time during the flight, the total energy of the ball must be same.

total energy = potential energy + kinetic energy

As the height of the object increases, the potential energy increases. Therefore the kinetic energy must decrease to keep the total energy constant

Answer 11

Oh i see
thanks @ParthKohli n @ganeshie8 :)

Answer 12

So I should draw them as an arch or is there actual values to observe?

Answer 13

@Somy

Answer 14

@Nnesha

Answer 15

@peachpi

Answer 16

@ParthKohli

Answer 17

@whpalmer4

Answer 18

@aaronq

Answer 19

I don't really understand this stuff.

Answer 20

yes, there are actual values that you can work out, and you have the information you need

Answer 21

But I don't understand what to do

Answer 22

Not to worry, I'll show you how to work it.

there's a general formula you can use for finding the height of an object if air resistance is ignored.

\[h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0\]where \(g\) is the acceleration due to gravity (here \(g = 10\text{ m/s}^2\))
\(v_0\) is the initial velocity of the object, with a positive value indicating upward travel
\(h_0\) is the initial height of the object, with a positive value indicating it is above the ground

We know that at \(t=0\) the value of \(h(t) = h(0) = 0\)
We also know that at \(t = 2\) the value of \(h(t) = h(2) = 0\)

The formula is that of an inverted parabola (the coefficient of the squared term is negative, so it opens downward), and hopefully you know that a parabola is symmetrical about the vertex and that the \(x\) coordinate of the vertex is given by \[x_{vertex} = -\frac{b}{2a}\]if we write the parabola in the form\[y = a x^2 + b x + c\]

Answer 23

What do we use the height for?

Answer 24

The height is a function of time, given a specific set of initial velocity, acceleration due to gravity, and initial height. We are given the acceleration due to gravity (10 m/s^2) and the initial height (0 m) but not the initial velocity. However, we also know 2 values of \(t\) at which the height is 0 m, so from that we can work out the initial velocity that makes the path of the object fit the constraints.

Put another way, we know the value of \(y\) at \(x = 0\) and \(x = 2\), and we know the values of \(a\) and \(c\), all we need to do is plug in some numbers and we can find \(b\).
Agreed?

Answer 25

So it's basically a^2 + b^2 =c^2?

Answer 26

a=0
c=2
Is that right?

Answer 27

No. No Pythagorean theorem here. Remember, the form of a parabola that opens up or down is
\[y = a x^2 + b x + c\]
compare that with our height function
\[h(t) = -\frac{1}{2}g t^2 + v_0 t + h_0\]

can you tell me what \(a,b,c\) are?

Answer 28

We are looking for b right?

Answer 29

We need all 3 of them, but yes, b is the most unknown :-)

Answer 30

a=10 m/s^2
c=20m

Answer 31

\[a = -\frac{1}{2}g\]\[b = v_0\]\[c = h_0\]

Answer 32

I don't understand why the v0 would be or how to get the height

Answer 33

\(v_0\) is the initial velocity in the vertical direction of our object in motion

Answer 34

a=5

Answer 35

*a=5 m/s^2

Answer 36

we can get the height at any time \(t\) for which our model is valid(*) by simply plugging \(t\) into the equation for \(h(t)\) and evaluating it. To get a numeric value, we will of course have to know all of the values of \(g, v_0, h, t\) in numeric form.

Close, but no cigar. This is pretty important, so I want you to take another look at what I wrote and get the correct answer.

Answer 37

(*) the model only holds true for the time when the object is launched until it lands. times before or after that period will give heights that do not correspond with our scenario (negative numbers). We are modeling the height with a portion of a parabola, but only the portion of the parabola that models the height is of interest — the rest of it does not apply.

Answer 38

a=-5m/s^2

Answer 39

You can think of that height function as three components:

the first component, with the \(t^2\) term, represents motion of the object due to gravity. It starts out at 0, and rapidly increases in magnitude, subtracting from the total height.

the second component, with the \(t\) term, represents motion of the object due to its initial velocity. It starts out at 0 and increases linearly with time, adding to the total height if the initial velocity is positive.

the third component, with the \(h_0\) term, is the initial position of the object.

If the object has no initial velocity (for example, dropping an object from a ladder), the height is a function only of the acceleration of gravity and the initial height.

if the object has no acceleration due to gravity, the height is a function only of the initial velocity and the initial height

Answer 40

Yes, that is correct. And what is the value of \(c = h_0\)?

Answer 41

Okay you're saying we don't know the v0 for this?

Answer 42

h(t)=1/2gt^2 +v0t + h0
h(2)=-5 +v0(2) +h0

Answer 43

Is that the formula I should be using to find h0?

Answer 44

we do not know \(v_0\) from the problem statement, but we can determine it from what we are given.

The only way to get \(h_0\) here is to read the problem. The ball departs from the ground, right? That means \(h_0 = 0\)

Answer 45

Oh. Okay

Answer 46

h(2)=-5 +v0(2) +0

Answer 47

Would that be right?

Answer 48

yes, but we also know the value of \(h(2)\)...what is it?

Answer 49

h(0)=-5 +v0(2) +0

Answer 50

sorry, not quite right:

\[h(2) = -5(2)^2 + v_0(2) + 0\]