AP Physics

Unit 1 - Vectors and Kinematics

Beginner

Mathematical

FRQ

You're a Phy Pro Member

Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
0

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

Step Derivation/Formula Reasoning
1 \Delta x = v_i t + \frac{1}{2} a t^2 Use the kinematic equation to find the time t , where \Delta x is the displacement, v_i is the initial velocity, and a is the acceleration due to gravity.
2 200 = 23t + \frac{1}{2}(9.8)t^2 Substitute the known values: \Delta x = 200 \, \text{m}, v_i = 23 \, \text{m/s}, and a = 9.8 \, \text{m/s}^2.
3 200 = 23t + 4.9t^2 Simplify the equation by computing \frac{1}{2} \times 9.8.
4 4.9t^2 + 23t – 200 = 0 Rearrange the equation into the standard quadratic form at^2 + bt + c = 0 .
5 t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} Use the quadratic formula (or use a graph) to solve for t .
6 t = \frac{-23 \pm \sqrt{23^2 – 4(4.9)(-200)}}{2(4.9)} Substitute a = 4.9 , b = 23, and c = -200 into the quadratic formula.
7 t = \frac{-23 + 66.7}{9.8} Select the positive root because time cannot be negative.
8 t \approx 4.45 \, \text{s} Solve for t . The object takes approximately 4.45 seconds to hit the ground.

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

Step Derivation/Formula Reasoning
1 v_f = v_i + at Use the kinematic equation to find the final velocity v_f , where v_i is the initial velocity, and a is the acceleration due to gravity.
2 v_f = 23 + 9.8 \times 4.45 Substitute the known values: v_i = 23 \, \text{m/s}, a = 9.8 \, \text{m/s}^2, and time t \approx 4.45 \, \text{s}.
3 v_f \approx 66.7 \, \text{m/s} Combine the terms to find the final velocity. The final velocity is approximately 66.6 m/s.

Need Help? Ask Phy To Explain This Problem

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

Simple Chat Box
NEW Smart Actions

Topics in this question

See how Others Did on this question | Coming Soon

Discussion Threads

Leave a Reply

  1. 4.46 seconds
  2. 66.7 m/s

Nerd Notes

Discover the world's best Physics resources

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.

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
Made By Nerd-Notes.com
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)}
\omega (Angular Velocity)\text{radians per second (rad/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

Error Report

Sign in before submitting feedback.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.