a) 2.75 s
The vertical position of the ball at time t is given by the equation
[tex]y= h+ut-\frac{1}{2}gt^2[/tex]
where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:
[tex]0 = 4 + 12t - 4.9 t^2[/tex]
This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:
[tex]v=u-gt[/tex]
where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:
[tex]v=12 -(9.8)(2.75)=-15.0 m/s[/tex]
And the negative sign means the direction is downward.
