Question 254699
A ball thrown in the air with a velocity of 24.6 m/s is at a height represented by h(t)=24.6t-4.9t^2 + 2 meters after t seconds.
=======
Note: h(t) is the height of the ball after t seconds.
=======
a) After how many seconds will the ball fall to the ground?
The height is zero when the ball is on the ground.
Solve -4.9t^2+24.6t+2 = 0
---
I graphed it to get:
t = 5.10 seconds 
------------------------------
b) When is the ball increasing? Decreasing?
The vertex is at t = -b/2a = -24.6/(2(-4.9)) = 2.51 seconds
Increasing: 0<= t <= 2.51 seconds
Decreasing: 2.51< t < 5.10 seconds
-----------------------------------------
c) After how many seconds is the height of the ball the greatest?
That's what the vertex tell you: t = 2.51 seconds
===================================================
Cheers,
Stan H.