Q#74
You stand at the top of a cliff while your friend stands on the ground below you. You drop a ball from rest and see that she catches it 1.4 s later. Your friend then throws the ball up to you, such that it just comes to rest in your hand. What is the speed with which your friend threw the ball?
Answer:
As shown in Example 2–15, the speed with which the ball was thrown upward is the same as its speed on returning to the ground.
From the symmetry of the two motions (both motions have speed = 0 at top, have same distance traveled, and have same acceleration), the time for the ball to rise is the same as the time for the ball to fall, 1.4 s.
Choose upward to be the positive direction and the origin to be at the level where the ball was thrown. For the ball, $v_y$ = 0 at the top of the motion, and a = −g.
Find the initial velocity from Eq. $v_y=v_{0y}+a_yt$.
$v_{0y}=v_y-a_yt$
$v_{0y}=0-(-9.80 \ m/s^2)(1.4 \ s)=13.72 \ m/s$
Q#75
On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 $\mu$m. A CD player’s readout laser scans along the spiral’s sequence of bits at a constant speed of about 1.2 m/s as the CD spins.
(a) Determine the number N of digital bits that a CD player reads every second.
(b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits, and so you might expect the required bit rate for a CD player to be
$N_0=2\left(44,100 \frac{samplings}{s}\right)\left(\frac{16 \ bits}{sampling}\right)=1.4 \times 10^6 \frac{bits}{s}$
where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that $N_0$ is less than the number N of bits (=$N-N_0$) actually read per second by a CD player.
The excess number of bits is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error-correction?
Answer:
(a) Multiply the reading rate times the bit density to find the bit reading rate.
$N = \frac{1.2 \ m}{s}\times \frac{1 \ bit}{0.28 \times 10^{-6} \ m}=4.3 \times 10^6 \ bits/s$
(b) The number of excess bits is $N-N_0$
$N-N_0=4.3 \times 10^6 \ bits/s-1.4 \times 10^6 \ bits/s=2.9 \times 10^6 \ bits/s$
$\frac{N-N_0}{N}=\frac{2.9 \times 10^6 \ bits/s}{4.3 \times 10^6 \ bits/s}=0.67 = 67$%
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