Q#46
(II) A helicopter is ascending vertically with a speed of 5.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is for the package?]
Answer:
Choose downward to be the positive direction, and take $y_0$ = 0 to be the height where the object was released.
The initial velocity is $v_{0y}$ = −5.40 m/s, the acceleration is a = 9.80 $m/s^2$ , and the displacement of the package will be y = 105 m.
The time to reach the ground can be found from Eq.
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
with x replaced by y.
$105 \ m=0+(-5.40 \ m/s)t+\frac{1}{2}(9.80 \ m/s^2)t^2$
$\frac{2 \times 105 \ m}{9.80 \ m/s^2}=\frac{(2 \times -5.40 \ m/s)}{9.80 \ m/s^2}t+t^2$
$t^2-1.102t-21.43=0$
$t_{12}=\frac{1.102 \pm \sqrt{(-1.102)^2-4(1)(-21.43)}}{2(1)}$
$t_{12}=\frac{1.102 \pm \sqrt{86.934}}{2}$
$t_{1}=\frac{1.102 +\sqrt{9.324}}{2}$ or $t_{2}=\frac{1.102 - \sqrt{9.324}}{2}$
Q#47
(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window. Roger’s room is on the third floor, 15 m above the sidewalk.
(a) How fast are the balloons traveling when they pass Roger’s window?
(b) Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is 5.0 m high.
Answer:
(a) Choose y = 0 to be the ground level and positive to be upward.
Then $y_0$ = 15 m, a = −g , and t = 0.83 s describe the motion of the balloon.
Use Eq. $y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$ so
$0=15 \ m+v_{0y}(0.83 \ s)+\frac{1}{2}(-9.80 \ m/s^2)(0.83 \ s)^2$
$0=v_{0y}(0.83 \ s)+11.624 \ m$
$v_{0y}=\frac{-11.642 \ m}{0.83 \ s}=-14.0 \ m/s$
(b) Consider the change in velocity from being released to being at Roger’s room, using Eq.
$v_y^2=v_{0y}^2+2a_y \Delta y$
$0=(-14.0 \ m/s)^2+2(-9.80 \ m/s) \Delta y$
$\Delta y = \frac{-196}{19.6}=-10.0 \ m$
Thus the balloons are coming from two floors above Roger, or the fifth floor.
Q#48
(II) Suppose you adjust your garden hose nozzle for a fast stream of water. You point the nozzle vertically upward at a height of 1.8 m above the ground (Fig. 2–40). When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.5 s. What is the water speed as it leaves the nozzle?
Answer:
Choose upward to be the positive direction and $y_0$ = 0 to be the location of the nozzle.
The initial velocity is $v_{0y}$, the acceleration is a = −9.80 $m/s^2$, the final location is y = −1.8 m, and the time of flight is t = 2.5 s.
Using Eq. $y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
and substituting y for x gives the following:
$-1.8 \ m=0+v_{0y}(2.5 \ s)+\frac{1}{2}(-9.80 \ m/s)(2.5 \ s)^2$
$-1.8 \ m=v_{0y}(2.5 \ s)-30.625 \ m$
$v_{0y}=\frac{28.825 \ m}{2.5 \ s}=11.53 \ m/s$
Q#49
(III) A falling stone takes 0.31 s to travel past a window 2.2 m tall (Fig. 2–41). From what height above the top of the window did the stone fall?
Answer:
Choose downward to be the positive direction and $y_0$ = 0 to be the height from which the stone is dropped.
Call the location of the top of the window $y_w$ , and the time for the stone to fall from release to the top of the window is $t_w$ .
Since the stone is dropped from rest, using Eq.
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$ (*)
with y substituting for x, we have
$y_w=y_0+v_{0y}t+\frac{1}{2}a_yt^2=0+0+\frac{1}{2}gt_w^2$
The location of the bottom of the window is $y_w$ + 2.2 m, and the time for the stone to fall from release to the bottom of the window is $t_w$ + 0.31 s.
Since the stone is dropped from rest, using Eq. (*), we have the following:
$y_w+2.2 \ m=y_0+v_{0y}t+\frac{1}{2}a_yt^2=0+0+\frac{1}{2}g(t_w+0.31 \ s)^2$
Substitute the first expression for yw into the second one and solve for the time.
$y_w+2.2 \ m=0+0+\frac{1}{2}g(t_w+0.31 \ s)^2$
$\frac{1}{2}gt_w^2+2.2 \ m=\frac{1}{2}g(t_w+0.31 \ s)^2$
$\frac{1}{2}gt_w^2+2.2 \ m=\frac{1}{2}g(t_w^2+2t_w (0.31)+(0.31)^2)$
$2.2=gt_w (0.31)+\frac{1}{2}g(0.31)^2$
$gt_w (0.31)= 2.2-\frac{1}{2}g(0.31)^2$
$t_w=\frac{2.2-\frac{1}{2}g(0.31)^2}{g(0.31)}=0.569 \ s$
Use this time in the first equation to find the desired distance.
$y_w=\frac{1}{2}gt_w^2=\frac{1}{2}(9.80 \ m/s^2)(0.569 \ s)^2=1.587 \ s$
Q#50
(III) A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.4 s later. If the speed of sound is how high is the cliff?
Answer:
For the falling rock, choose downward to be the positive direction and $y_0$ = 0 to be the height from which the stone is dropped. The initial velocity is 0 υ = 0 m/s, the acceleration is a = g, the final position is y = H, and the time of fall is $t_1$. Using Eq.
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
with y substituting for x, we have
$H=0+0+\frac{1}{2}gt_1^2$
For the sound wave, use the constant speed equation that
$v_s=\frac{\Delta x}{\Delta t}=\frac{H}{T-t_1}$
which can be rearranged to give
$t_1=T-\frac{H}{v_s}$
where T = 3.4 s is the total time elapsed from dropping the rock to hearing the sound. Insert this expression for $t_1$ into the equation for H from the stone, and solve for H.
$H=\frac{1}{2}g\left(T-\frac{H}{v_s}\right)^2$
$\frac{g}{2v_s^2}H^2-\left(\frac{gT}{v_s}-1\right)+\frac{1}{2}gt^2=0$
$4.239 \times 10^{-5}H^2-1.098 H+56.64=0$, so
$H = 51.7 \ m$ or $H = 2.59 \times 10^4 \ m$
If the larger answer is used in $t_1=T-\frac{H}{v_s}$ a negative time of fall results, so the physically correct answer is 51.7 m
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