Why is initial velocity confusing during free fall?

[moomoocow]

initial velocity confusion! please help!

hello all,
i am slightly confused about initial velocity during free fall
if a ball is being thrown in the air, and the velocity time graph shows 20m/s at 0 seconds, and -20m/s at 4 seconds, which is when the ball drops
if i want to calculate the displacement, i know that i can use the formula d=(vo)(t) +0.5(a)(t^2)
with vo being initial velocity
why is it that i would have to put 0 for vo? why isn't it 20m/s?

 

[neutrino]

Which displacement are you talking about? (I'm assuming that the ball is thrown vertically)

 

[oksanav]

I'm not sure I understand your question. If a ball is thrown with initial velocity of 20 m/s, then that's your initial velocity. The V0 just means velocity at t=0. If a ball is dropped or released then its V0 is 0.

 

[moomoocow]

the ball is thrown vertically, displacement being the change in distance. oksanav: why is it that If a ball is dropped or released then its V0 is 0?, but then why would the v-t graph show 20m/s at 0 seconds?

 

[neutrino]

When the ball's thrown initially, it has a velocity of 20 m/s, but as it reaches the top, its velocity goes to zero. After that, its velocity increases, now pointing down, reaching 20m/s just before hitting the ground. Depending on which part you're calculating, you're initial velocity will change.

 

[radou]

the ball is thrown vertically, displacement being the change in distance. oksanav: why is it that If a ball is dropped or released then its V0 is 0?, but then why would the v-t graph show 20m/s at 0 seconds?

If a ball is dropped, there usually is no initial velocity, but there can be some. But in that case, it is a throw. It's all about directions.