How high does it go before returning to the ground?

Question

A grapefruit is tossed straight up with an initial velocity of 54 ft/sec. The grapefruit is feet 5 above the ground when it is released. Its height at time t  is given by y=-16t^2+54t+5.

Answer 1

?????????? "initial velocity"???????????

yu alredee told up it go strate up, so now yu need tu giv the SPEED

????? "initial" ?????????...its the start speed or speed at time=0

start speed=54 ft/sek

gravity slo it bi 32.2 ft/sek-sek

time tu get tu the top hite=54/32.2=1.770 seks

aveaej speed=(1/2)(start speed +end speed)=(1/2)(54+0)=27

vertikal distans it go=27*1.770=47.79 ft

Wen it come down, gotta go xtra 5 ft tu hit ground, so vertikal drop=52.79 ft

drop start speed=0, so speed down=g*t

then distans=(1/2)g*t*t=52.79 ft

so t=root(2*52.79/g)=root(105.58/32.2)=root(3.27882

=1.811 seks

hit ground with speed=32.2*1.811=58.3 ft/sek