Question

I'll give 99 points

Answer 1

Maximum height is at dH/dt=0

H= -16T^2+80T+5

dH/dt=-32T+80

-32T+80=0

32T=80

T=2.5s when height is max

Hmax=-16(2.5)^2+80*2.5)+5

=105ft

catching the ball at 5' means solve for T at H=5

H= -16T^2+80T+5=5

-16T^2+80T=0

-16T(T-5)=0

so T=0 cannot be the ans or T=5s

Answer 2

H= -16T^2+80T+5
so its vertex is at T=b/2a=80/(2*(-16))=2.5
H(2.5)=-16*2.5^2+80*2.5+5=105
last one
H=5
-16T^2+80T+5=5
-16T^2+80T=0
T=0 when ball is thrown or T=5 when it is caught