## Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Explanation 0 likes
0

# (a) Rock’s Speed Just Before It Falls Into the River

Step Derivation/Formula Reasoning
1 v_i = 20 \, \text{m/s} Initial velocity of the rock thrown upward.
2 h = -42 \, \text{m} Final displacement of the rock from the point of throw to the river is -42 meters (negative since it’s downward).
3 v_f^2 = v_i^2 + 2a h Using the kinematic equation to find the final velocity, where v_i is initial velocity, v_f is final velocity, a is acceleration and h is displacement.
4 a = -9.8 \, \text{m/s}^2 Acceleration due to gravity (negative since it’s downward).
5 v_f^2 = (20 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)(-42 \, \text{m}) Substitute the known values into the kinematic equation.
6 v_f^2 = 400 + 823.2 Calculate the expression inside the equation.
7 v_f = \sqrt{1223.2} \, \text{m/s} Taking the square root of both sides to solve for the final velocity v_f.
8 v_f \approx 35 \, \text{m/s} Final speed of the rock just before it falls into the river.

# (b) Time Taken for the Rock to Strike the River

Step Derivation/Formula Reasoning
1 v_f = v_i + a t Using the first kinematic equation to solve for time t, where v_f is final velocity, v_i is initial velocity, a is acceleration, and t is time.
2 v_f = -35 \, \text{m/s} Since the rock is falling downward, take final velocity v_f as negative.
3 -35 \, \text{m/s} = 20 \, \text{m/s} + (-9.8 \, \text{m/s}^2) t Substitute the known values into the time equation.
4 -35 = 20 – 9.8 t Simplify the equation.
5 -55 = -9.8 t Combine like terms.
6 t = \frac{55}{9.8} Isolate t by dividing both sides of the equation by -9.8.
7 t \approx 5.6\, \text{s} Solving for time gives approximately 5.6 seconds.

## Need Help? Ask Phy To Explain This Problem

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

Simple Chat Box

1. 35.2 m/s
2. 5.6 seconds

## 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