SOLUTION: The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the a
Question 111036: The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the air, how many seconds have passed until the ball hits the ground??
I really don't understand how to put this problem in the equations it's supposed to go into and I don't understand how to solve them...help! Found 2 solutions by scott8148, ankor@dixie-net.com:Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the air, how many seconds have passed until the ball hits the ground??
:
Look at the three elements of the equation:
-16t^2 is the force of gravity, negative because it is pulling it down
:
+32t is the speed of the ball upward when it is thrown
:
+3 is the point above the ground that the ball thrown
:
t is in seconds
:
h = height in feet
:
When the ball hits the ground the height is 0, so we have:
;
-16t^2 + 32t + 3 = 0
:
Use the quadratic formula to find t; a = -16; b = 32; c = 3
:
:
:
:
: ; we only want this solution
:
:
t = +2.089 sec, when it hits the ground
:
A graph would look like this. Vertical = height; horizontal = time in sec
:
Note that it crosses the y axis at 3' when t = 0 and max height is about 19 ft at 1 sec
;
Did this help you understand this?
