# Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.

1. (a) What is the difference in the time the balls spend in the air? (2 points)
2. (b) What is the velocity of each ball as it strikes the ground? (2 points)
3. (c) How far apart are the balls 0.800 s after they are thrown? (3 points)
1. Difference in Time Spent in the Air: \boxed{\Delta t = 0.5 , \text{seconds}}
2. Velocity of Each Ball as it Strikes the Ground:
• Ball Thrown Downward: \boxed{v_{\text{down}} = 53.9 , \text{m/s}}
• Ball Thrown Upward: \boxed{v_{\text{up}} = 19.6 , \text{m/s}}
3. Distance Apart after 0.800 Seconds: \boxed{\text{Distance Apart} = 2.352 , \text{meters}}
0
1. Difference in Time Spent in the Air:
Step Formula Derivation Reasoning
1 H = v_0t_{\text{down}} + \frac{1}{2}gt_{\text{down}}^2 Kinematic equation for downward motion; solve for t_{\text{down}
2 t_{\text{up-to-balcony}} = \frac{v_0}{g} Time to reach the highest point and back to the balcony height for upward motion
3 H = \frac{1}{2}gt_{\text{up-additional}}^2 Additional time from balcony to ground for upward motion
4 \Delta t = t_{\text{down}} – t_{\text{up}} Difference in time spent in the air
5 \boxed{\Delta t = 0.5 , \text{seconds}} Final Calculation
2. Velocity of Each Ball as it Strikes the Ground:
Ball Thrown Downward Ball Thrown Upward Reasoning
v_{\text{down}} = v_0 + gt_{\text{down}} v_{\text{up}} = \sqrt{2gH} Velocity calculation at impact for both balls
\boxed{v_{\text{down}} = 53.9 , \text{m/s}} \boxed{v_{\text{up}} = 19.6 , \text{m/s}} Final combination
3. Distance Apart after 0.800 Seconds:
Ball Thrown Downward Ball Thrown Upward Reasoning
y_{\text{down}} = v_0t – \frac{1}{2}gt^2 y_{\text{up}} = H – (v_0t – \frac{1}{2}gt^2) Position calculation after 0.800 s; distance apart
\boxed{\text{Distance Apart} = 2.352 , \text{meters}}

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1. Difference in Time Spent in the Air: \boxed{\Delta t = 0.5 , \text{seconds}}
2. Velocity of Each Ball as it Strikes the Ground:
• Ball Thrown Downward: \boxed{v_{\text{down}} = 53.9 , \text{m/s}}
• Ball Thrown Upward: \boxed{v_{\text{up}} = 19.6 , \text{m/s}}
3. Distance Apart after 0.800 Seconds: \boxed{\text{Distance Apart} = 2.352 , \text{meters}}

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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