Answer
(a) At t = 10.0 s, the rocket is a height of 467 meters above the Earth's surface.
(b) The speed is 110 m/s when the rocket is 325 meters above the Earth's surface.
Work Step by Step
$a_y(t) = (2.80~m/s^3)~t$
Note that $v_{0y} = 0$ and $y_0 = 0$ because the rocket is initially on the ground at rest.
$v_y(t) = v_{0y}+ \int_{0}^{t}a_y(t)~dt$
$v_y(t) = 0 + \int_{0}^{t}(2.80~m/s^3)~t~dt$
$v_y(t) = (1.40~m/s^3)~t^2$
We can use $v_y(t)$ to find $y(t)$.
$y(t) = y_0+ \int_{0}^{t}v_y(t)~dt$
$y(t) = 0+ \int_{0}^{t} (1.40~m/s^3)~t^2~dt$
$y(t) = \frac{1.40~m/s^3}{3}~t^3$
(a) At t = 10.0 s,
$y = \frac{1.40~m/s^3}{3}~(10.0~s)^3 = 467~m$
At t = 10.0 s, the rocket is a height of 467 meters above the Earth's surface.
(b) We can find the time $t$ when y = 325 m.
$y(t) = \frac{1.40~m/s^3}{3}~t^3 = 325~m$
$t^3 = \frac{(325~m)(3)}{1.40~m/s^3}$
$t = 8.864~s$
At t = 8.864 s,
$v_y = (1.40~m/s^3)~(8.864~s)^2$
$v_y = 110~m/s$
The speed is 110 m/s when the rocket is 325 meters above the Earth's surface.
