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A ball is kicked into the air and follows the path described by h(t)= – 4.9t^2 + 6t + 0.6, where t is the time in seconds and h is the height in meters above the ground. Find the maximum height of the ball. What value would you have to change in the equation if the maximum height of the ball is more than 2.4 meters?

Verified Answer :

apologiabiology

I'm going to do the calculus way because it's easier
max height,
take derititive of the equation

h'(t)=-9.8t+6
find what value of t makes it equal to 0
t=6/9.8
if we inputed it for t we get
h(6/9.8)=2.43673...

2.43673...>2.4 so we don't need to change anything

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