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3 years ago

I'll give 99 points

Mathematics

2 answers:

photoshop1234 [79]3 years ago

8 0

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

Leto [7]3 years ago

3 0

<span>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
</span>-16T^2+80T+5=5
-16T^2+80T=0
T=0 when ball is thrown or T=5 when it is caught

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3 years ago

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2 years ago

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anyanavicka [17]

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Read 2 more answers

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nadezda [96]

Answer:

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3 years ago