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# Helllllppppp !! :(((

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A baseball is hit so that it travels straight upward after being struck by the bat. If the ball takes 4.5 s to reach its maximum height, then the ball's initial velocity is

Sep 22, 2020

#1
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A baseball is hit so that it travels straight upward after being struck by the bat. If the ball takes 4.5 s to reach its maximum height, then the ball's initial velocity is

Hello Guest!

$$h=v_0\cdot t-\frac{1}{2}gt^2\\ v_0\cdot t=\frac{1}{2}gt^2\\ v_0= \frac{1}{2}gt\\ v_0=\frac{1}{2}\cdot9.81\frac{m}{sec^2}\cdot4.5sec$$     That is the wrong approach. sorry.

$$v_0=22.0725m/sec$$

!

Sep 22, 2020
edited by asinus  Sep 22, 2020
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Asinus, this is NOT the correct equation for this question.

--. .-

GingerAle  Sep 22, 2020
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v = vo - 1/2 at      v = 0    a = 9.8 m/s2       t = 4.5 s

0  = v - 1/2 (9.8)(4.5)

vo = 1/2 (9.8)(4.5) = 22.05 m/s           (Just like asinus found !)

Sep 22, 2020
edited by ElectricPavlov  Sep 22, 2020
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vo = 1/2 (9.8)(4.5) = 22.05 m/s           (Just like asinus found !)

Yes, and this is demonstrable evidence that dumbness is contagious.

This equation will work if the total time of flight is used.  4.5 seconds is half   (½) of the total flight time.

Or take the direct approach and use the equation Alan presented, below ...

--. .-

GingerAle  Sep 22, 2020
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Agree with Alan..... I made the same mistake Asinus did

v = v- at        where   v = 0     v0 = initial velocity      a = 9.81 m/s^2       t = 4/5 sec        (there is no 1/2 in the equation!!)

Oops !    THANX , Alan !

ElectricPavlov  Sep 22, 2020
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I think it is more an example of the power of suggestion .....

I do not think either answer is 'dumb'......just incorrect because of a mistake....

ElectricPavlov  Sep 22, 2020
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I think it is more an example of the power of suggestion ....

Ok... But why didn’t the suggestion of “...this is NOT the correct equation for this question,” have any power?

...Id est, why did you believe that re-solving the equation would confirm its validity?

I do not think either answer is 'dumb'......just incorrect because of a mistake...

I do not think the answers are dumb either, but I do think ...dumbness is contagious.

--. .-

GingerAle  Sep 22, 2020
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The correct equation is $$v = v_0 + at$$

Here $$v = 0m/s; a = -9.81m/s^2;t=4.5s$$

Sep 22, 2020
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Hello ElectricPavlov, hello Alan!

Thanks for the correction.

I'm trying differential calculus. Is that OK?

$$\color{blue}h(v)=vt-\frac{v^2}{2g}\\ \frac{dh(v)}{dv}=t-\frac{v_0}{g}=0\\ v_0=gt=\frac{9.81m\cdot4.5s}{s^2}$$

$$v_0=44.145m/s$$

$$h_{max}=v_0\cdot t-\frac{v_0^2}{2g}\\ h_{max}=44.145m/s\cdot4.5s-\frac{44.145^2}{2\cdot 9.81}m$$

$$h_{max}=99.33m$$

!

Sep 22, 2020
edited by asinus  Sep 22, 2020