- #1

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1) What is the initial speed v0 of the ball?

2)What is the height of the ball when the vertical speed is 1/2 of the initial speed vo?

Thanks

(I'm so sorry for my bad english. It's not my mother tongue)

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- Thread starter Spruance
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- #1

- 33

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1) What is the initial speed v0 of the ball?

2)What is the height of the ball when the vertical speed is 1/2 of the initial speed vo?

Thanks

(I'm so sorry for my bad english. It's not my mother tongue)

Last edited:

- #2

Astronuc

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Well, v_{o} is the magnitude of the velocity, and since velocity is a vector, there must be a unit vector associated with that magnitude, however . . .

one must determine the vertical displacement after the ball is deflected up, and actually, I believe the appropriate question is "what is the height of the ball when the vertical speed is 1/2 of the initial speed v_{o}?"

One could use conservation of energy.

one must determine the vertical displacement after the ball is deflected up, and actually, I believe the appropriate question is "what is the height of the ball when the vertical speed is 1/2 of the initial speed v

One could use conservation of energy.

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- #3

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0.5mv^2+mgh = 0.5mv0^2+mgh0

v0^2 = -2mgh0 + mv^2 + 2mgh

v0 = (-2gh0 + v^2 + 2gh)^1/2

v0 = (- 2 * 9.81 * 4 + o m/s + 2 * 9.81 * 3)^1/2

v0 = (- 78 + 58.86)^1/2

Here is something wrong

- #4

Astronuc

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One is solving for the h, when v = v

0.5 m (v

0.5 (v

h = h

- #5

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Maybe I must edit my first question.

What I meant was:

1) How fast must the initial speed v0 be, if the ball shall reach 4 meters into the air (at the heighest, of course)

- #6

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I tried to rearrange your equation:

0.5 (vo/2)2 + gh = 0.5 vo2 + gho

I get:

h = ((v0^2)-(v0^8)+gh0))/(g)

Maybe I am just stupid

- #7

Astronuc

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v

v

and lets not forget the 1/2 in front of the v

so the equation for h should be

h = 1/2 (3/4 v

3/8 (v

- #8

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v0^2 = v^2 + 2gh - gh0

v0 = (v^2 + 2gh - gh0)^1/2

v0 = (2g(h-h0))^1/2

v0 = (2 * 9.81 (4 m -3 m))^1/2

v0 = 4.43 m/s

What is wrong?

- #9

nrqed

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you forgot a 2 in the first equation but then you put it back later so the answer should be right. How do you know it's not? (if you are using Webassign, it may be simply a question of sig figs)Spruance said:

v0^2 = v^2 + 2gh - gh0

v0 = (v^2 + 2gh - gh0)^1/2

v0 = (2g(h-h0))^1/2

v0 = (2 * 9.81 (4 m -3 m))^1/2

v0 = 4.43 m/s

What is wrong?

- #10

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Let us assume that the initial speed is 4.43 m/s

then the height of ball is given:

h = 1/2 (3/4 vo2)/g + ho =

3/8 (vo2)/g + ho

= 3/8 * ((4.43^2)/9.81) + 3

= 3.75 m

Is that correct?

- #11

nrqed

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you shoudl have mentioned that you has switched to question 2 now...Spruance said:

Let us assume that the initial speed is 4.43 m/s

then the height of ball is given:

h = 1/2 (3/4 vo2)/g + ho =

3/8 (vo2)/g + ho

= 3/8 * ((4.43^2)/9.81) + 3

= 3.75 m

Is that correct?

Yes, that looks right.

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