given
d(t)=-16t^2-4t+382
1. times at which rock is 310 from ground
solve d(t)=310=-16t^2-4t+382
=>
16t^2+4t-72=0
t=(-4+/-sqrt(4^2+4*16*72)/(2*16)
=> t=-9/4 or t=2
Reject negative root from context, so
time=2 seconds.
2.
maximum is when d'(t)=0, or
-32t-4=0 => t=-1/8
Therefore rock was descending since t=0 (can also see from -4t, i.e. initial velocity is -4 ft/s)
Rock reaches ground when d(t)=0, i.e.
-16t^2-4t+382=0
t=-5.01 or t=4.763 seconds. reject negative root.
Therefore, domain, t=0 to 4.763 seconds
range = 0 to 382 ft
