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
+15001 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.
\(v_0=22.0725m/sec\)
!
+37159 v = vo - 1/2 at v = 0 a = 9.8 m/s2 t = 4.5 s
0 = v o - 1/2 (9.8)(4.5)
vo = 1/2 (9.8)(4.5) = 22.05 m/s (Just like asinus found !)
+2490 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 ...
--. .-
+37159 Agree with Alan..... I made the same mistake Asinus did
v = vo - 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 !
+37159 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....
+2490 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.
--. .-
+33659 The correct equation is \(v = v_0 + at\)
Here \(v = 0m/s; a = -9.81m/s^2;t=4.5s \)
+15001 Hello ElectricPavlov, hello Alan!
Thanks for the correction.
I'm trying differential calculus. Is that OK?
\(\color{blue}h(v)=vt-\frac{v^2}{2g}\\ \frac{dh(v)}{dv}=t-\frac{v_0}{g}=0\\ v_0=gt=\frac{9.81m\cdot4.5s}{s^2}\)
\(v_0=44.145m/s\)
\(h_{max}=v_0\cdot t-\frac{v_0^2}{2g}\\ h_{max}=44.145m/s\cdot4.5s-\frac{44.145^2}{2\cdot 9.81}m\)
\(h_{max}=99.33m\)
!
