# Problems with Velocity and Acceleration

1. Feb 22, 2006

### physhelp

A ball thrown straigh upward from the ground reaches a maximum height of 400ft. What was its initial velocity? Using the formulas:
V=at+Vo
X=1/2at^2+Vot+Xo
the distance travelled is 400ft, and the initial distance was 0. The force of gravity is -32ft/s/s. Also, the velocity of the ball at the top of the height is 0.

in the equation V=at+Vo, I don't know how to solve it with 2 variables (t and Vo)

Last edited: Feb 22, 2006
2. Feb 22, 2006

### jollygood

V=at+Vo ---------------(A)
X=1/2at^2+Vot+Xo -----------(B)

ok. couple of things to note

assume we are measuring height from ground. (i.e. X0 = 0)
at maximum height, the velocity of the ball becomes zero for a moment just before it turns around. so at the point V = 0

so equation (A) becomes
0 = at + V0 => t = - V0/a

here 'a' is gravitational acceleration. u need to put correct sign as well.

i.e. a = -(g). that way your time is positve. (time cannot be negative)

now substitute this in to (B) and solve for V0.

3. Feb 22, 2006

### physhelp

thanks but I'm still stucki got Vo^2=12800/-15 and you can't do a square root of a negative :S

4. Feb 22, 2006

### physhelp

thank you so much, i got it. except i made Vo=32t instead. :D