# Equation for initial velocity

Problem:
A ball is thrown vertically upward, which is the positive direction. A little while later it returns to its point of release. The ball is in the air for a total time of 10 s. Note: Near the earth's surface, g is approximately 9.80 m /s2.

(a1) What is the algebraic expression for the initial velocity v0 of the ball? Express your answer in terms of the ball's displacement y, its acceleration a in the vertical direction, and the elapsed time t.

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is this correct?:
Equation:
v0=(y-.5at^2)/t

v0=(0-490)/10
v0=(-490)/10
v0=-49

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tony873004
Gold Member
"A ball is thrown vertically uwpard which is the positive direction"

And you get -49, which since its negative, is the down direction. Will a ball thrown towards the ground last 10 seconds in the air? Only if you throw it from the roof of a tall building.

You seem to have used the formula y=ut+0.5a(t^2), transposed it and obtained u=(y/t)-0.5at, so as long as you have your positive and negative directions right you've done it right, i think.

one thing to think about, if the positive direction is upwards, how could the ball travel up when it has a negative initial velocity?

Homework Helper
The vertical 'throw' equation for the y-direction is $$y(t)\vec{j}=v_{0}t\vec{j}-0.5gt^2\vec{j}$$, which is equal to y(t) = v0*t - 0.5*g*t^2. Actually, I don't see any problems with the directions; the initial velocity v0 is 'positive', and gravitiy is acting in the 'negative' direction all the time. I don't see how you got a negative velocity from your equation. v0 equals 49 [m/s].

i got negative because, the displacement is 0,
v0=(y-.5at^2)/t
v0=(0-.5(9.8)(10^2))/10
v0=(0-490)/10
v0=-490/10
v0=-49

However when i use the equation radou gives, i get the positive 49... so i think its just the eqution i used..... Thanks guys!

nrqed
Homework Helper
Gold Member
Joules23 said:
i got negative because, the displacement is 0,
v0=(y-.5at^2)/t
v0=(0-.5(9.8)(10^2))/10
v0=(0-490)/10
v0=-490/10
v0=-49

However when i use the equation radou gives, i get the positive 49... so i think its just the eqution i used..... Thanks guys!
Your equation is correct, it's just that the acceleration due to gravity is $a_y = -g = -9.80 m/s^2$. (assuming the positive y axis pointing upward). You used a=+g which is the problem.

Hope this helps