Question

Can someone help me with a physics question? There's no one in the Physics room.

Answer 1

it says 85 online :P

Answer 2

You toss a ball straight up with an initial speed of 30 m/s. How high does it go and how long is it in the air? (neglecting air resistance)

Answer 3

yeah but no one is responding

Answer 4

45.92 meters.. i think

Answer 5

i just need to find out how high it goes, since i already found out it will be in the air for 6 seconds.

Answer 6

how did you find that out

Answer 7

Do you know the formula

\[h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0\]?

Answer 8

whpalmer got it :P gonna leave

Answer 9

umm what does the Vot and the Ho mean ? is it initial?

Answer 10

\[g = 9.8 \frac{\text{m}}{\text{s}^2} = 32\frac{\text{ft}}{\text{s}^2} \]\[v_0 = \text{initial velocity}\]\[h_0=\text{initial height}\]

Answer 11

the equation has the form of a parabola, and so the value of \(t\) where \(h(t)=0\) will be the time of flight, and the vertex of the parabola will be the highest point in the flight.

Answer 12

oh okay thanks

Answer 13

initial height here is 0 m (well, really it'd be whatever height the ball left your hand, but whatever). initial velocity is 30 m/s. Set h(t) = 0 and solve for the positive-valued solution to get the time of flight. Do you remember how to find the vertex of a parabola?

Answer 14

i dont :x

Answer 15

are you just useing that equation you sent me earlier?

Answer 16

no problem. If you have a parabola in the standard form
\[y = ax^2+bx+c\](which we do)
the x-value of the vertex can be found by \(x = -\dfrac{b}{2a}\). That will give you the exact value of \(t\) at which the ball is at its peak height. Then you plug that value of \(t\) into \(h(t)\) to find the peak height. Easy, right?

Answer 17

Yes, that's the equation to use.
\[h(t) =- \frac{1}{2}9.8t^2 +30t = -4.9t^2+30t\]

Answer 18

is 9.8 given?

Answer 19

it's the value of \(g\) on the Earth's surface, to 2 significant figures.

Answer 20

then what is -b/2a?

Answer 21

http://www.wolframalpha.com/input/?i=g+unit&a=UnitClash_*g+unit.*StandardAccelerationOfGravity--

Answer 22

is it -30/2(4.9)?

Answer 23

Yes, that's almost correct.

\[h(t) = -4.9t^2+30t (+ 0)\]\[y=ax^2+bx+c\]
matching up bits and pieces
\[a=-4.9,\, b = 30, \,c = 0\]

vertex is at \(x = -\dfrac{b}{2a}\) or \(t = -\dfrac{30}{2*(-4.9)} = \)

Answer 24

im putting it in my calculator but its not coming out right

Answer 25

what is the answer suppose to be?

Answer 26

what are you getting?

Answer 27

3.06

Answer 28

Okay. What's wrong with that? You understand that that is the value of \(t\) when the ball is at the top of its flight, right? Here's a graph of height (y value) vs time (x value)

Answer 29

so thats the answer?

Answer 30

That is the time at which the ball is at the peak. Now you need to find the height at that time, by evaluating the formula for \(h(t) = h(3.06) = -4.9t^2+30t = \)

Answer 31

ok now i got 46

Answer 32

i think i get it now
thanks

Answer 33

Now what's the total time of flight?

Answer 34

yeah i think so.

Answer 35

46 meters

Answer 36

no, that's the maximum height. How long is the ball in the air? Set \(h(t) = 0\) and solve for \(t\). You'll get two solutions, one of which is \(t=0\).

Answer 37

6 seconds right?

Answer 38

6.12 seconds. I don't know if you're using significant figures or not here.

Answer 39

the time of flight will be exactly twice the time of max height in the case where we have the ball starting and ending on the ground, because parabolas are symmetric about the vertex.

Answer 40

idk ): my physics prof is just giving me formulas to go off of, he doesnt really use algebra

Answer 41

That's no reason why you can't apply knowledge you have.

Answer 42

work smarter, not harder :-)

Answer 43

yeah, but the problem is ive already forgotten everything i learned before, its been awhile since ive taken any math courses

Answer 44

okay. well, for doing these problems, you need to know how to find the vertex of a parabola. have you had calculus?

Answer 45

would you know how to find the answer from d=vt?

Answer 46

find which answer?

Answer 47

the answer to the problem we just did

Answer 48

after finding out 6 seconds in the time it takes

Answer 49

using d=vt to find out how high it goes?

Answer 50

how high did the ball go? no, we can't do that, because there's gravity accelerating the ball downward. If the ball didn't get acted upon by gravity or other forces, then you could use that formula to find the height of the ball at any given time.

if you look at the h(t) formula, there's the first term \((-(1/2) gt^2)\) which represents the displacement due to gravitational acceleration. the second term \((v_0 t)\) represents the displacement due to the initial velocity, and the third term (which was 0\) is the displacement due to the initial position. At the end of the ball's path, the first two terms are equal in magnitude but opposite in sign. So you could do the problem like this:
\[4.9t^2 = 30t\]Factor out a \(t\)\[4.9t = 30\]\[t = \frac{30}{4.9} = 6.122\]

Then if you observed that it is a parabola, and the starting and ending heights are equal, you can use the symmetry of a parabola to determine that the vertex is exactly halfway through the flight. However, if the ball was thrown off a building (so \(h_0>0)\), and landed on the ground \(h = 0\), we couldn't use that shortcut because the parabola won't be symmetrical about the vertex, as seen in the attached diagram.

Answer 51

the vertical line is at the halfway point, the horizontal line crosses through the vertex, and you can see that they do not intersect in the vertex (as they would if the ball was launched from the ground).

Answer 52

ok thanks

Answer 53

dang so long 0_0 nice one whpalmer

Answer 54

I like my dead horses well-beaten :-)

Answer 55

so, remember this formula:
\[h(t) = -\frac{1}{2}gt^2+v_0t+h_0\]the vertex will be at \[t_{max} = \frac{v_0}{g}\]just plug in the value in the first equation to get the maximum height:\[h(t_{max}) = -\frac{1}{2}g(\frac{v_0}{g})^2+v_0(\frac{v_0}{g})+h_0\]if you want it all written out, but normally you'd compute the value of \(t_{max}\) and then plug that in.

Answer 56

That last thing simplifies to \[h(t_{max}) = \frac{v_0^2}{2g}+h_0\]which is probably worth remembering if you do a lot of these problems, although I never remember it and get along fine...

In this problem,
\[h(t_{max}) = \frac{30^2}{2(9.8)} + 0 = \frac{900}{19.6} = 45.9184 \]