A ball is thrown directly upward from height 6 ft with initial velocity of 20 ft/sec. The function #s(t) = -16t^2 + 20t + 6# gives height of the ball t seconds after thrown. Determine the time where the ball reaches its max height and find the max height?

1 Answer
Shwetank Mauria
Oct 24, 2017

The ball reaches its maximum height of #12 1/4# feet after #5/8# seconds.

Explanation:

As at #t=0#, #s(t)=6# and as #v(t)=s'(t)=-32t+20#, the initial velocity #v(0)=20#. Hence these conditions are built within the function. As such we can proceed with the function to find maximum height.

The function #s(t)=-16t^2+20t+6# gives its maxima when converted to vertex form. So doing so

#s(t)=-16t^2+20t+6#

= #-16(t^2+20/16t)+6#

= #-16(t^2+5/4t)+6#

= #-16(t^2+2xx(5/8)t+(5/8)^2)+16(5/8)^2+6#

= #-16(t-5/8)^2+25/4+6#

= #-16(t-5/8)^2+6 1/4+6#

= #-16(t-5/8)^2+12 1/4#

Hence, the ball reaches its maximum height of #12 1/4# feet after #5/8# seconds.

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