Inactive Tutor answered 01/15/26
v(.59)=h'=0=-9.8t+vo
,59= vo/9.8
v0=9.8(.59)=5.782
h= -4.9t^2 +vot +ho
h(t) =-4.9t^2 +5.782t +1.6
max height= h(5.9)= -4.9(5.9)^2+5.782(5.9)+1.6=3.30569 not 3.25 m but very close
use a calculator
Amanda H.
asked 12/06/18A student estimated based on the video that the ball left my hand 1.6 meters off the ground, and after 0.59 seconds the ball reached the maximum height of 3.25 meters.
Use this information to find an equation for the height, h of the ball, in meters, after t seconds.
Inactive Tutor answered 01/15/26
v(.59)=h'=0=-9.8t+vo
,59= vo/9.8
v0=9.8(.59)=5.782
h= -4.9t^2 +vot +ho
h(t) =-4.9t^2 +5.782t +1.6
max height= h(5.9)= -4.9(5.9)^2+5.782(5.9)+1.6=3.30569 not 3.25 m but very close
use a calculator
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