AP Physics Unit

Unit 1 - Vectors and Kinematics

Advanced

Proportional Analysis

GQ

In the absence of air resistance, a projectile is launched from, returns to ground level and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

92

Projectile motion range is derived as follows:

  1. Express the horizontal and vertical components of the initial velocity.
  2. Determine the time of flight based on vertical motion.
  3. Calculate the horizontal distance traveled in this time.
Step Formula Derivation Reasoning
1.1 v_{x} = v \cos(\theta) Horizontal component of initial velocity, constant since no air resistance.
1.2 v_{y0} = v \sin(\theta) Initial vertical component of the velocity.
2.1 t = \frac{2v_{y0}}{g} Time of flight, derived from v_{y} = v_{y0} – gt and the fact that at maximum height, v_{y} = 0, hence t = \frac{2v_{y0}}{g} for a round trip.
2.2 t = \frac{2v \sin(\theta)}{g} Substitute v_{y0} = v \sin(\theta).
3.1 R = v_{x} t Range is the horizontal distance traveled.
3.2 R = v \cos(\theta) \times \frac{2v \sin(\theta)}{g} Substitute v_{x} = v \cos(\theta) and t from step 2.2.
3.3 R = \frac{v^2 \sin(2\theta)}{g} Simplify using trigonometric identity \sin(2\theta) = 2 \sin(\theta) \cos(\theta).

Now we apply this to find the new range when the launch speed is doubled.

Step Formula Derivation Reasoning
4.1 R’ = \frac{(2v)^2 \sin(2\theta)}{g} New range formula with doubled speed 2v.
4.2 R’ = 4 \times \frac{v^2 \sin(2\theta)}{g} Simplify to show the relationship with the original range.
4.3 R’ = 4R Substitute the original range R.

Given the original range of 23 m, we will calculate the new range.

The new range, calculated from the derived formula with the launch speed doubled, is 92.0 , \text{m}.

Topics in this question

Discover how students preformed on this question | Coming Soon

Discussion Threads

Leave a Reply

92

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing, you agree to the updated Terms of Sale, Terms of Use, and Privacy Policy.

Suggest an Edit

What would you like us to add, improve, or change on this page? We listen to all your feedback!

Nerd-Notes.com
KinematicsForces
\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2F = m \cdot a
v = v_i + a \cdot tF_g = \frac{G \cdot m_1 \cdot m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu \cdot N
R = \frac{v_i^2 \cdot \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{m \cdot v^2}{r}KE = \frac{1}{2} m \cdot v^2
a_c = \frac{v^2}{r}PE = m \cdot g \cdot h
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m \cdot v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum m \cdot r^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k \cdot 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}
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 (kg·m/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (N·m)}
I (Moment of Inertia)\text{kilogram meter squared (kg·m}^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!
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.

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

NEW!

Join Elite Memberships and get 25% off 1-to-1 Elite Tutoring!