Question

A ball is thrown into the air from a height of 4ft . The height,h , of the ball after t seconds, is given by the equation h=4.9t^2+20t+4 . What is the maximum height the ball reaches?

Answer 1

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