Answer
(a) $2.64~s$
(b) The amount of time the person is above a height of $14.0~ft$ is more than the amount of time the person is below a height of $14.0~ft$.
(c) $T_{below}=1.86~s$ and $T_{above}=0.78~s$
Work Step by Step
$1~ft=0.305~m$
$28.0~ft=(28.0~ft)(\frac{0.305~m}{1~ft})=8.54~m$
(a) Let the positive direction be upward with origin at the blanket.
First, let's find the initial velocity. Remember that, at the maximum height, the speed is zero.
$v=0$, $\Delta x=8.54~m$, $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=v_0^2+2(-9.81~m/s^2)(8.54~m)$
$v_0=±\sqrt {2(9.81~m/s^2)(8.54~m)}$
$v_0=±12.94~m/s$. When a person is thrown, she moves upward, at first, so:
$v_0=12.94~m/s$
The time spent by the person in the air ($T$) is the double of the time it takes to reach the maximum height ($t$).
$v=0$, $v_0=12.94~m/s$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$0=12.94~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=12.94~m/s$
$t=\frac{12.94~m/s}{9.81~m/s^2}$
$T=2t=2\frac{12.94~m/s}{9.81~m/s^2}=2.64~s$
(b) As the person moves upward, her speed decreases. So, when the person is above $14.0~ft$ her speed is less than her speed when she is below $14.0~ft$. Hence, the amount of time the person is above a height of $14.0~ft$ is more than the amount of time the person is below a height of $14.0~ft$.
(c) First, let's find the amount of time the person is below a height of $14.0~ft$ ($T_{below}$). We need to find the time it takes for the person to reach $14.0~ft$ ($t_{below}$), that is, from $x=0$ to $x=14.0~ft$. This time is equal to the time it takes from $x=14.0~ft$ to $x=0$, now moving downward. So, $T_{below}=2t_{below}$.
$14.0~ft=(14.0~ft)(\frac{0.305~m}{1~ft})=4.27~m$
$\Delta x=4.27~m$, $v_0=12.94~m/s$, $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$v^2=(12.94~m/s)^2+2(-9.81~m/s^2)(4.27~m)$
$v=±\sqrt {(12.94~m/s)^2+2(-9.81~m/s^2)(4.27~m)}$. The person is moving upward at first.
$v=9.147~m/s$
$v=v_0+at$
$9.147~m/s=12.94~m/s+(-9.81~m/s^2)t_{below}$
$(9.81~m/s^2)t_{below}=9.147~m/s$
$t_{below}=\frac{9.147~m/s}{(9.81~m/s^2)}$
$T_{below}=2t_{below}=2\frac{9.147~m/s}{(9.81~m/s^2)}=1.86~s$
$T_{above}=T-T_{below}=2.64~s-1.86~s=0.78~s$
