Q#71
Two students are asked to find the height of a particular building using a barometer. Instead of using the barometer as an altitude measuring device, they take it to the roof of the building and drop it off, timing its fall. One student reports a fall time of 2.0 s, and the other, 2.3 s. What % difference does the 0.3 s make for the estimates of the building’s height?
Answer:
Choose downward to be the positive direction and the origin to be at the top of the building. The barometer has $y_0$ = 0, $v_{0y}$ = 0, and a = g = 9.80 $m/s^2$.
Use Eq. $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$ to find the height of the building, with x replaced by y.
$y-0=0+\frac{1}{2}(9.80 \ m/s^2)t^2$
$y_{t=2.0 \ s}=\frac{1}{2}(9.80 \ m/s^2)(2.0 \ s)^2=19.6 \ m$
$y_{t=2.3 \ s}=\frac{1}{2}(9.80 \ m/s^2)(2.3 \ s)^2=25.9 \ m$
The difference in the estimates is 6.3 m. If we assume the height of the building is the average of the two measurements, then the % difference in the two values is
$\frac{6.3 \ m}{22.75 \ m} \times 100=27.7$%
The intent of the method was probably to use the change in air pressure between the ground level and the top of the building to find the height of the building.
The very small difference in time measurements, which could be due to human reaction time, makes a 6.3-m difference in the height. This could be as much as 2 floors in error.
Q#72
Two children are playing on two trampolines. The first child bounces up one-and-a-half times higher than the second child. The initial speed up of the second child is 4.0 m/s.
(a) Find the maximum height the second child reaches.
(b) What is the initial speed of the first child?
(c) How long was the first child in the air?
Answer:
Assume that $y_0$ = 0 for each child is the level at which the child loses contact with the trampoline surface. Choose upward to be the positive direction.
(a) The second child has $v_{02y}$ = 4.0 m/s, a = g = −9.80 $m/s^2$, and v = 0 m/s at the maximum height position.
Find the child’s maximum height from Eq. $v_y^2=v_{02y}^2+2a(y_2-y_0)$, with x replaced by y.
$y_2=y_0+\frac{0-(4.0 \ m/s)^2}{2(-9.80 \ m/s^2)}=0.8163 \ m$
(b) Since the first child can bounce up to one-and-a-half times higher than the second child, the first child can bounce up to a height of 1.5(0.8163 m) = 1.224 m = $y_1-y_0$.
$v_y^2=v_{01y}^2+2a_y(y-y_0)$ is again used to find the initial speed of the first child.
$v_{01y}=\sqrt{v_y^2-2a_y(y-y_0)}$
$v_{01y}=\sqrt{0-2(-9.80 \ m/s^2)(1.224 \ m)}=4.898 \ m/s$
The positive root was chosen since the child was initially moving upward.
(c) To find the time that the first child was in the air, use Eq. $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$ with a total displacement of 0, since the child returns to the original position.
$y-y_0=v_{01y}t_1+\frac{1}{2}a_yt_1^2$
$0=(4.898 \ m/s)t_1+\frac{1}{2}(-9.80 \ m/s^2)t_1^2$
$0=(4.898 \ m/s)-4.90 \ m/s^2 t_1)t_1$
so, $t_1=0$ or $t_1=\frac{4.898}{4.90}\approx 1.0 \ s$
The time of 0 s corresponds to the time the child started the jump, so the correct answer is 1.0 s .
Q#73
If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3200 m to an altitude of 450 m, where she will open her parachute? What would her speed be at 450 m? (In reality, the air resistance will restrict her speed to perhaps )
Answer:
Choose downward to be the positive direction and the origin to be at the location of the plane. The parachutist has $v_{0y}$ = 0, a = g = 9.80 $m/s^2$, and will have $y-y_0$ = 3200 m – 450 m = 2750 m when she pulls the ripcord. Eq. $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$, with x replaced by y, is used to find the time when she pulls the ripcord.
$y-y_0=0+\frac{1}{2}a_yt^2$
$t=\sqrt{\frac{2(y-y_0)}{a_y}}$
$t=\sqrt{\frac{2(2750 \ m)}{9.80 \ m/s^2}}=23.69 \ s$
The speed is found from Eq.
$v_y=v_{0y}+a_yt$
$v_y=0+(9.80 \ m/s^2)(23.69 \ s)=232.16 \ m/s$ or
$v_y=(232.16 \ m/s)\left(\frac{3.6 \ km/h}{1 \ m/s}\right) = 840 \ km/h$
This is well over 500 miles per hour!
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