SOLUTION: A ball is thrown upward with an initial speed of 24.5m/s. When is it 19.6m high? (Two answers) I dont get one part of it or even how to start it

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A ball is thrown upward with an initial speed of 24.5m/s. When is it 19.6m high? (Two answers) I dont get one part of it or even how to start it      Log On


   

Question 327445: A ball is thrown upward with an initial speed of 24.5m/s. When is it 19.6m high? (Two answers) I dont get one part of it or even how to start it
Answer by galactus(183) About Me  (Show Source):
You can put this solution on YOUR website!
The height can be found by using
y=v%2At-1%2F2%2Ag%2At%5E2
Where v=24.5 and g=9.8 m/s^2 (the gravity constant in metric)
All we have to do is set y=19.5 and solve for t:
19.5=24.5t-4.9t%5E2
-4.9t%5E2%2B24.5t-19.5=0
Solve the quadratic and we see t=.99 and t=4
The ball is 19.5 m going up and again when it comes back down.
It takes .99 seconds to reach 19.5 m on the way up and 4 seconds to descend back down to 19.5m on its way down.
We can also find how high it goes and how long it takes to get there.
The max height can be found by using t=-b%2F%282a%29
t=-24.5%2F%282%2A%28-4.9%29%29=2.5
It takes 2.5 seconds to reach its max height.
Plug this back into our quadratic to find that max height:
-4.9%282.5%29%5E2%2B24.5%282.5%29=30.625 meters.
It reaches a max height of a little over 30 m.
I know this is more than you needed, but it may come in handy.:):)