解此题？

from the question,we got
initial velocity Vo = 224ft/s upward
height H = 528ft
gravity acceleration g = 9.8m/s^2 = 32.2ft/s^2 downward
from the formular H = Vot- 1/2gt^2
we get the equation: 224t - 16.1t^2 = 528
solve equation,we get two answers: 3.01s or 10.9s
but the rocket is not like a ball,it won't come back,
and the answer is the less one
so after 3.01 sec the rocket reach a height of 528ft.

assume after t seconds the rocket reaches a height of 528 ft, THEN:

(224-g*t+224)*t/2=528, here g=9.8 m/s^2,

solve that for t, we get:

t=[448-sqrt(448^2-1056*4g)]/2g, plug in g=9.8, we get

t=2.5 s

...这是物理题嘛
224t-1/2*9.8*t^2=528
解方程的t