- #1

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(a) How high does it go?

? m

(b) How long is it in the air?

? s

How do I solve this

- Thread starter thschica
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- #1

- 47

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(a) How high does it go?

? m

(b) How long is it in the air?

? s

How do I solve this

- #2

HallsofIvy

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v(t)= v

where v(t) is the velocity of the ball at time t seconds after the ball was hit staight up and v

h(t)= h

where h(t) is height of the ball at time t seconds after the ball is hit and h

As long as the velocity is positive, the ball is still going up- not yet at its highest point.

When the velocity is negative, the ball is coming back down- already past its highest point.

What do you think the velocity is

Put that into the equation and solve for t to find when the ball is at its highest point. Now use that time in the height equation to find that height.

"How long is it in the air?"

It's in the air until it gets back to the ground! What is its height when it hits the ground? If you put that into the height equation and solve for t what do you get?

- #3

TD

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Just a small note, HallsofIvy probably meant g = 9.8 m/s² (the 1 probably came from 9.8**1**).

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- #5

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Nevermind the answer is 4.5 seconds right?

- #6

mezarashi

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How high does it go?

Vinitial = 22m/s, a= gravity = -9.8m/s/s, Vfinal = 0 m/s (at its peak)

d = ?

How long does it take to get to the ground? Using the conservation of energy the velocity when it reaches the ground again must be equal to when it went up, Vfinal = -22m/s, a = -9.8m/s/s, Vinitial=22m/s, t=?

- #7

HallsofIvy

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The answer tothschica said:Nevermind the answer is 4.5 seconds right?

As far as "I cant find the velocity of the ball at its hightest point." I was trying to give you that: as long as the velocity is positive, the ball is still going up and so is not at its highest point! If the velocity is negative, the ball is now going down and so is past its highest point! The velocity at the highest point can't be positive or negative so it must be 0.

Yes, I meant to write that g= -9.81. Don't know where the "8" got to!

If v(t)= 22- 9.81t= 0, then 9.81t= 22 so t= 22/9.81= 2.24 s when the ball is at its highest point.

Now use h(t)= 22t- 4.91t

Of course the ball ends its flight when h(t)= 0 again. You can solve

22t- 4.91t

22t- 4.91t

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