Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Explanation 0 likes
0

# (a) Determine the time it takes for the object to hit the ground.

Step Derivation/Formula Reasoning
1 d = v_0t + \frac{1}{2}gt^2 Use the kinematic equation for distance fallen under gravity. Note initial velocity is 0 and therefore is removed in the next step.
2 t = \sqrt{\frac{2d}{g}} Rearrange the formula to solve for time ( t ).
3 t = \sqrt{\frac{2 \times 45}{9.8}} Substitute d = 45 \, \text{m} and g = 9.8 \, \text{m/s}^2 into the equation.
4 t = \sqrt{\frac{90}{9.8}} Simplify the division inside the square root.
5 t \approx 3.03 \, \text{s} Take the square root to find the time. This is the final result.

# (b) Determine the velocity of the object when it hits the ground.

Step Derivation/Formula Reasoning
1 v_f = v_i + at \rightarrow v_f = gt Use the kinematic equation for final velocity. Note that initial velocity is 0 and acceleration is due to gravity.
2 v = 9.8 \, \text{m/s}^2 \times 3.03 \, \text{s} Substitute g = 9.8 \, \text{m/s}^2 and the time from part (a) t \approx 3.03 \, \text{s} .
3 v \approx 29.7 \, \text{m/s} Multiply to find the final velocity. This is the final result.

# (c) How long does it take for the object to fall 30 m?

Step Derivation/Formula Reasoning
1 d = \frac{1}{2}gt^2 Use the kinematic equation for distance fallen under gravity (0 initial velocity) as done in part (a).
2 t = \sqrt{\frac{2d}{g}} Rearrange the formula to solve for time ( t ).
3 t = \sqrt{\frac{2 \times 30}{9.8}} Substitute d = 30 \, \text{m} and g = 9.8 \, \text{m/s}^2 into the equation.
4 t \approx 2.47 \, \text{s} Final result.

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

1. 3.03 seconds
2. 29.7 m/s
3. 2.47 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