Answer:
a) 625
b) Yes
c) No
Step-by-step explanation:
a)To calculate the maximum height we can use the derivative to find the slope of the function.
[tex]dh/dt=-32t+200[/tex]
To find the slope we make the equation equal to 0:
[tex]-32t+200=0[/tex]
[tex]t=200/32[/tex]
The slope at 0 is t=200/32=6.25[/tex]
The maximum height at t=6.25 is
[tex]h=-16(6.25^2)+200(6.25)=625[/tex]
The maximum height is 625
b) at t=8
[tex]-16(8^2)+200(8)=576[/tex]
The height will be 576 which is less than 625
The two x-intercepts are when h=0:
c)
[tex]-16(14^2)+200(14)=-336[/tex]
[tex]-16t^2+200t=0[/tex]
[tex]-2t^2+25t=0[/tex]
[tex]t(-2t+25)=0[/tex]
[tex]t=0[/tex]
[tex]t=12.5[/tex]
The ball is not in the air because the ball will only be in the air from t=0 to t=12.5
