Question 886833
h(t) = -16t^2 +(vo)(t) + ho, where vo=velocity in ft/sec and ho= initial height in ft
we are given vo = 30 and ho = 4, therefore
h(t) = -16t^2 +30t +4
now we know that the x coordinate of the max altitude is -b/2a and in our problem
t = -b/2a = -30/(2*-16) = 0.9375
so max height is reached in 0.9375 seconds, now we can calculate the max height
h(0.9375) = -16*(0.9375^2) +30*0.9375 +4
h(0.9375) = 18.0625 feet
The flyer's center of gravity does not reach 20 feet
now find initial velocity to reach 25 feet
25 = -16*(0.9375^2) +vo*0.9375 +4
25 = −14.0625 +vo*0.9375 +4
vo*0.9375 = 35.0625
vo = 37.4 feet per second
therefore the initial velocity must be 37.4 feet per second to reach a height of 25 feet