- #1

diego1404

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If a ball is thrown upward to find out the height of a flag by timing when the ball passes the flag how do you find the height given that the ball passes the flag twice. When only time is given. So lets say t=5seconds and t=15 seconds

I know how to solve it but i dont understand the concept

I will use #### to ask questions along the way

What i would do is use the formula

x=xi +vi + (1/2)(-g)(t^2)

lets say Equation 1 is

x=vi(5)-4/9(.5^2)

then subtract t2-t1 and divide it by 2 so: (t2-t1)/2

this would give us time 2.5

we can get velocity when time is 2.5s by vf-va=a*t so, 0-va=-9.81(2.5)

Va=24.525

###why cant i go initial time t=5 seconds we can get velocity by a*t =vf-vi where vi=0

Vf-vi= -49.05, vi=49.05

then substitute x with equation 1 earlier and va=24.25 then find out vi

(va^2)-(vi^2)= 2gx

###why is Va(final velocity) not zero here if Vf(final velocity) was zero earlier when time was 2.5 seconds

(24.525^2)-(vi^2)=(2)(9.81)((vi(5)-4/9(.5^2))

then you use quadratic equation to find vi

plug in vi to the initial equation to get distance

#### another question is why is initial velocity needed to find out the distance

#### why is vi not zero since it starts at the bottom

#### if it an object was dropped above a certain height its vi is zero

#### what i was thinking as my main solution was that x=(-1/2)(9.81)(.5^2)

############### why was that wrong

what i dont understand is that if this motion was horizontal ie. a wheel moving in the ground you can have a vi=0 but in the question above vi cannot equal zero.

i always hear that distance is a function of time, which i dont understand

mathematically you can plug in zero but why doesnt it make sense here>