here is the method:
http://www.jiskha.com/display.cgi?id=1391459618
You throw a ball straight into the air from a height of 4 feet and with a speed of 8.22 m/s. The moment the ball leaves your hand you start running away at a speed of 3.36 m/s. How far are you from the ball, the moment it hits the ground?
(what are the formulas used?)
5 answers
how did you find t? my physics professor is not the greatest at explaining so I am extremely lost and don't know at all how to do this.
I solved the quadratic equation
if
a x^2 + b x + c = 0
then
x = [-b +/- sqrt (b^2-4ac) ]/2a
You had this in algebra I suspect.
It works fine for
0 = (Hi-h) + Vi t - (1/2) g t^2
if
a x^2 + b x + c = 0
then
x = [-b +/- sqrt (b^2-4ac) ]/2a
You had this in algebra I suspect.
It works fine for
0 = (Hi-h) + Vi t - (1/2) g t^2
as with any parabola, it crosses the finish height twice, once on the way up and again later on the way down. We want it later coming back down so use the + sqrt and not the - sqrt for t
okay, now I get it. thank you for your help!
1 answer