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

Guest Sep 22, 2020

#1**+1 **

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.

\(v_0=22.0725m/sec\)

!

asinus Sep 22, 2020

#3**+1 **

v = v_{o} - 1/2 at v = 0 a = 9.8 m/s^{2 }t = 4.5 s

0 = v _{o }- 1/2 (9.8)(4.5)

v_{o} = 1/2 (9.8)(4.5) = 22.05 m/s (Just like asinus found !)

ElectricPavlov Sep 22, 2020

#5**0 **

*v*_{o}* = 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

#6**+1 **

Agree with Alan..... I made the same mistake Asinus did

v = v_{o }- at where v = 0 v_{0} = 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

#7**+1 **

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

#8**0 **

*I think it is more an example of the power of suggestion ....*

Ok... But why didn’t the ** suggestion** of “...

...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

#4**+5 **

The correct equation is \(v = v_0 + at\)

Here \(v = 0m/s; a = -9.81m/s^2;t=4.5s \)

Alan Sep 22, 2020

#9**+1 **

**Hello ElectricPavlov, hello Alan!**

Thanks for the correction.

I'm trying differential calculus. Is that OK?

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

\(v_0=44.145m/s\)

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

\(h_{max}=99.33m\)

!

asinus Sep 22, 2020