Answer
(a) By a factor of 2.
(b) By a factor of 4.
(c) $t_2=0.82~s=2\times0.41~s=2\times t_1$
$\Delta x_2=\frac{16.0~m^2/s^2}{19.62~m/s^2}=4\times\frac{4.0~m^2/s^2}{19.62~m/s^2}=4\times\Delta x_1$
Work Step by Step
(a) Let's find the time it takes to the youngster to reach the top of her flight when her speed will be zero. Also, you must remember that it takes the same time to the youngster to move back and reach the trampoline again.
Let the positive direction be upward with origin at the trampoline.
$a=-g=-9.81~m/s^2$
$v=v_0+at$
- initial speed = $2.0~m/s$:
$0=2.0~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=2.0~m/s$
$t=\frac{2.0~m/s}{9.81~m/s^2}$. But, it takes the same time to the youngster to move back and reach the trampoline again:
$t_1=2\times\frac{2.0~m/s}{9.81~m/s^2}=0.41~s$
- initial speed = $4.0~m/s$:
$0=4.0~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=4.0~m/s$
$t=\frac{4.0~m/s}{9.81~m/s^2}$. Again:
$t_2=2\times\frac{4.0~m/s}{9.81~m/s^2}=0.82~s$
We can conclude that: $t_2=2t_1$
(b) - initial speed = $2.0~m/s$:
$v^2=v_0^2+2a\Delta x_1$
$0^2=(2.0~m/s)^2+2(-9.81~m/s^2)\Delta x_1$
$(19.62~m/s^2)\Delta x_1=4.0~m^2/s^2$
$\Delta x_1=\frac{4.0~m^2/s^2}{19.62~m/s^2}=0.20~m$
- initial speed = $4.0~m/s$:
$v^2=v_0^2+2a\Delta x_2$
$0^2=(4.0~m/s)^2+2(-9.81~m/s^2)\Delta x_2$
$(19.62~m/s^2)\Delta x_2=16.0~m^2/s^2$
$\Delta x_2=\frac{16.0~m^2/s^2}{19.62~m/s^2}=0.82~m$
$\Delta x_2=4\Delta x_1$. Notice that: $0.82~m\ne4\times0.20~m$, but $\Delta x_2=\frac{16.0~m^2/s^2}{19.62~m/s^2}=4\times\frac{4.0~m^2/s^2}{19.62~m/s^2}=4\times\Delta x_1$
(c) See the items above.
