A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find the ball's initial velocity.

Question

lgbasallote · Answer

we use $x = V_{ox} t + \frac 12 at^2$ right?

lgbasallote · Answer

and a would be -9.8 m/s^2 right?

lgbasallote · Answer

wait doesn't sound right...

shubhamsrg · Answer

why dont you go for d(x)/dt = 0 => for maximum height
a=-9.8
t=3
u= ?

note that its the same thing as v-u = at
v=0

lgbasallote · Answer

but what will you take the derivative of? there's no equation

anonymous · Answer

the derivative of position is velocity.

so you get

v = at

anonymous · Answer

woops, v = at + vo

lgbasallote · Answer

wait..you guys are using $$V = V_o + at$$ 
right?

lgbasallote · Answer

so $$0 = V_o + (-9.8)(3)?$$

anonymous · Answer

yeah. thats the derivative of x = xo + vot + (1/2) a t^2

lgbasallote · Answer

lol never noticed that =))

lgbasallote · Answer

so is my equation right??

anonymous · Answer

yeah it is.

then go back to the position equation, use that Vo, you already know the t and the a. Just get the x.

lgbasallote · Answer

so to find the maximum height i use x = vot + 1/2 at^2 right?

anonymous · Answer

exactly!

lgbasallote · Answer

great! thanks :D