Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Explanation 0 likes
0

(a) How far does the textbook travel horizontally after it is released?

Step Derivation/Formula Reasoning
1 v_{0x} = v_0 \cos(\theta) Calculate the initial horizontal velocity. Use the initial speed and the angle of projection.
2 v_{0y} = v_0 \sin(\theta) Calculate the initial vertical velocity. Use the initial speed and the angle of projection.
3 v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s} Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the horizontal velocity formula.
4 v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s} Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the vertical velocity formula.
5 y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} Use the equation of motion in the vertical direction. The textbook is moving under gravity.
6 0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2 Set the displacement y to zero because we are calculating the time t when the textbook reaches the ground. g = 9.8 \text{m/s}^2 .
7 4.9 t^2 – 11.8 t – 12 = 0 Simplify the quadratic equation to solve for t .
8 t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} Use the quadratic formula where a = 4.9 , b = -11.8 , and c = -12 .
9 t \approx 3.18 \, \text{s} Solve the equation and take the positive root. This is the time the textbook stays in the air.
10 x = v_{0x} \cdot t Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time.
11 x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} Substitute v_{0x} = 16.2 \, \text{m/s} and t = 3.18 \, \text{s} into the horizontal distance formula to get the final answer.
\text{The horizontal distance traveled is approximately } 51.5 \, \text{m}

(b) What is the book’s velocity (speed and direction) when it reaches the ground?

Step Derivation/Formula Reasoning
1 v_y = v_{0y} – g t Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time.
2 v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s} Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , and t = 3.18 \, \text{s} into the vertical velocity formula.
3 v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} Calculate the magnitude of the total velocity using the Pythagorean theorem.
4 v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} Substitute v_{x} = 16.2 \, \text{m/s} and v_y = -19.4 \, \text{m/s} into the total velocity formula.
5 \theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) Calculate the direction of the velocity. Use the inverse tangent to find the angle.
6 \theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ The vector points 50.1° below the x-axis.
\text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ

(c) What is the book’s maximum height above the ground?

Step Derivation/Formula Reasoning
1 v_y = 0 The vertical velocity at the maximum height is zero.
2 v_y = v_{0y} – g t Use the vertical motion equation to find the time to reach maximum height.
3 0 = 11.8 – 9.8 t Set final vertical velocity v_y = 0 and solve for t .
4 t = \frac{11.8}{9.8} \approx 1.20 \, \text{s} Solving the equation gives the time to reach maximum height.
5 H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} Use the vertical motion equation to find the maximum height.
6 H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , t = 1.2 \, \text{s} , and initial height = 12 m.
7 H \approx 19.1 \, \text{m} Calculate the maximum height above the ground.
\text{The maximum height above the ground is approximately } 19.1 \, \text{m}

Need Help? Ask Phy To Explain This Problem

Phy can also check your working. Just snap a picture!

Simple Chat Box

See how Others Did on this question | Coming Soon

a) 51.5 m

b) 25.3 m/s at 50.1° below the horizontal

c) 19.1 m

Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Enjoying UBQ? Share the 🔗 with friends!

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!

Phy Pro

The most advanced version of Phy. Currently 50% off, for early supporters.

\$11.99

per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro