a.
the starting point is where t=0
d(0)=-16(0)^2+96(0)+112
D(0)=112
started from 112 feet
b. max height is vertex
for
y=ax^2+bx+c
the x value of the vertex is -b/2a
so
D(t)=-16t^2+96t+112
t value of vertex is -96/(2*-16)=3
it reaches after 3 seconds
C. simpliy evaluate D(t) for t=3
D(3)=-16(3)^2+96(3)+112
D(3)=-16(9)+288+112
D(3)=-144+400
D(3)=256
max height is 256ft
