Question

A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation that models the path of the ball is given by h = -4.9t^2 + 18t + 10. What is the maximum height of the ball?

Answer 1

Answer:

Yes it is very easy di I can't answer

Answer 2

Answer:

the answer is-

Step-by-step explanation:

The Equation of Height of the ball will be given by

H=h + ut + at²/2

Where h is the initial height =1.21metre

u = initial velocity

a = -g=-9.8ms^-2

[The equation you have written have a mistake, it is h=-4.9t^2 +20t+4, because  without negative sign,the ball will never stop at a height]

the equation of ball is given H= 4 + 20t + 4.9t^2

After comparing with above equation

u=20m/s and on maximum height v=0

But v is also the differentiation of height wrt time =-9.8t +20

so 0=-9.8t+20

t=20/9.8=2 (almost)s

So the height reached from 4feet =h=-4.9(2^2) +20(2) +4

=-18.6 +40+4

=25.4metre

toatal height =25.4 + 1.21=26.61

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