# An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane’s altitude is 732 m, the pilot releases a package.

1. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (3 points)
2. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (4 points)
1. 1379.29
2. 66.1°
0
Step Formula / Calculation Reasoning
1 v_{x} = v_{0} \cos(\theta), v_{y} = v_{0} \sin(\theta) Decomposing the initial velocity into horizontal (v_{x}) and vertical (v_{y}) components.
2 v_{x} = 97.5 \cos(50^\circ) \approx 62.68 , \text{m/s}, v_{y} = 97.5 \sin(50^\circ) \approx 74.67 , \text{m/s} Calculating the horizontal and vertical components of the initial velocity.
3 y = y_{0} + v_{y} \cdot t – \frac{1}{2} g t^{2} Kinematic equation for vertical motion, where y is the final height, y_{0} is the initial height, g is the acceleration due to gravity, and t is the time.
4 0 = 732 + 74.67 \cdot t – \frac{1}{2} \cdot 9.81 \cdot t^{2} Substituting values into the vertical motion equation to find the time when the package reaches the ground.
5 t \approx 11.80 s Solving the quadratic equation for time, taking the positive root as the physical solution.
6 x = v_{x} \cdot t Horizontal distance formula, where x is the distance and v_{x} is the constant horizontal velocity.
7 x = 62.68 \cdot 11.80 \approx 1379.29 m Calculating the horizontal distance from the release point to where the package hits the earth.
8 v_{y \text{ final}} = v_{y} – g \cdot t Calculating the final vertical velocity of the package just before impact.
9 v_{y \text{ final}} = 74.67 – 9.81 \cdot 11.80 \approx -36.05 m/s Finding the final vertical component of velocity.
10 \theta_{\text{impact}} = \arctan\left(\frac{v_{y \text{ final}}}{v_{x}}\right) Calculating the angle of the velocity vector with respect to the horizontal just before impact.
11 \theta_{\text{impact}} \approx -66.07^\circ Evaluating the angle of impact.

The distance along the ground from the point directly beneath the point of release to where the package hits the earth is approximately 1379.29 meters. The angle of the velocity vector of the package relative to the ground just before impact is approximately -66.07°.

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1. 1379.29
2. 66.1°

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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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