Answer
(a) The maximum height is 378 meters.
(b) Austin is 184 meters above the ground when the helicopter crashes into the ground.
Work Step by Step
(a) Let $y_1$ be the height reached by the helicopter while the engines are running.
$y_1 = \frac{1}{2}at^2 = \frac{1}{2}(5.0~m/s^2)(10.0~s)^2$
$y_1 = 250~m$
After 10.0 s,
$v_1 = at = (5.0~m/s^2)(10.0~s) = 50~m/s$
We can find the maximum height $y_2$ reached by the helicopter.
$y_2-y_1 = \frac{v_2^2-v_1^2}{2g} = \frac{0-(50~m/s)^2}{(2)(-9.80~m/s^2)}$
$y_2-y_1 = 128~m$
$y_2 = 128~m+250~m = 378~m$
The maximum height is 378 meters.
(b) After the engines are shut off, we can find the time $t$ for the helicopter to fall to the ground.
Let $v$ be the velocity when the helicopter hits the ground.
$v^2 = v_1^2 + 2a(y-y_0)$
$v = \sqrt{v_1^2 + 2a(y-y_0)}$
$v = \sqrt{(50~m/s)^2 + (2)(-9.80~m/s^2)(-250~m)}$
$v = -86~m/s$
Note that the negative sign means the velocity is directed toward the ground.
$t = \frac{v-v_1}{g} = \frac{-86~m/s-50~m/s}{-9.80~m/s^2} = 13.9~s$
We can find Austin's height $y_3$ 7.0 seconds after stepping out of the helicopter.
$y_3 = y_1+ v_1t + \frac{1}{2}gt^2$
$y_3 = 250~m+ (50~m/s)(7.0~s)+ \frac{1}{2}(-9.80~m/s^2)(7.0~s)^2$
$y_3 = 359.9~m$
We can find Austin's velocity $v_3$ 7.0 seconds after stepping out of the helicopter.
$v_3 = v_1 + at = 50~m/s - (9.80~m/s^2)(7.0~s)$
$v_3 = -18.6~m/s$
Austin continues falling for another 6.9 seconds with an acceleration of $-2.0~m/s^2$. We can find Austin's height $h$ at the moment when the helicopter reaches the ground.
$h = 359.9~m - (18.6~m/s)(6.9~s) - \frac{1}{2}(2.0~m/s^2)(6.9~s)^2$
$h = 184~m$
Austin is 184 meters above the ground when the helicopter crashes into the ground.
