Answer
(a) $y(t) = 450-4.9~t^2$
(b) It takes $9.58~s$ for the stone to reach the ground.
(c) The stone strikes the ground with a velocity of $93.9~m/s$ (downward).
(d) It takes $9.09~s$ for the stone to reach the ground.
Work Step by Step
(a) $a(t) = \frac{dv}{dt} = -9.8$
$v(t) = -9.8~t+C_1$
Since $v(0) = 0$, then $C_1 = 0$
$v(t) = \frac{dy}{dt} = -9.8~t$
$y(t) = -4.9~t^2+C_2$
Since $y(0) = 450,$ then $C_2 = 450$
$y(t) = 450-4.9~t^2$
(b) We can find the time $t$ when $y(t) = 0$:
$y(t) = 450-4.9~t^2 = 0$
$4.9~t^2 = 450$
$t^2 = \frac{450}{4.9}$
$t = \sqrt{\frac{450}{4.9}}$
$t = 9.58~s$
It takes $9.58~s$ for the stone to reach the ground.
(c) When $t = 9.58~s,$ then $v = (-9.8)(9.58) = -93.9~m/s$
The stone strikes the ground with a velocity of $93.9~m/s$ (downward).
(d) $v(t) = \frac{dy}{dt} = -5-9.8~t$
$y(t) = 450-5~t-4.9~t^2 = 0$
We can use the quadratic formula:
$t = \frac{5 \pm \sqrt{(-5)^2-(4)(-4.9)(450)}}{2(-4.9)}$
$t = \frac{5 \pm \sqrt{25+8820}}{-9.8}$
$t = -10.11, 9.09$
Since $t$ is positive, the solution is $t = 9.09~s$
It takes $9.09~s$ for the stone to reach the ground.