AP Physics

Unit 1 - Vectors and Kinematics

Advanced

Mathematical

FRQ

You're a Pro Member

Supercharge UBQ

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
0

Part (a): Time to wall

Derivation/Formula Reasoning
\[ \Delta x = v_i \cos 30^{\circ}\, t \] Horizontal motion: distance \( \Delta x = 45\,\text{m} \) equals horizontal speed times time.
\[ t = \frac{\Delta x}{v_i \cos 30^{\circ}} \] Solve previous relation algebraically for time \( t \).
\[ t = \frac{45}{29.4\cos 30^{\circ}} \approx 1.77\,\text{s} \] Insert the value of \( v_i \) found in part (b) to evaluate the time.
\[ \boxed{t \approx 1.77\,\text{s}} \] Time taken for the ball to reach the wall.

Part (b): Initial speed

Derivation/Formula Reasoning
\[ y = y_0 + v_i \sin 30^{\circ}\, t – \tfrac12 g t^{2} \] Vertical position of the ball; here \( y_0 = 1.5\,\text{m} \) and final \( y = 12.2\,\text{m} \).
\[ 12.2 = 1.5 + v_i \sin 30^{\circ}\, t – 4.9 t^{2} \] Substitute heights and \( g = 9.8\,\text{m/s}^2 \).
\[ 10.7 = 0.5 v_i t – 4.9 t^{2} \] Simplify numerical terms.
\[ t = \frac{45}{v_i \cos 30^{\circ}} \] Replace \( t \) using the horizontal relation from part (a).
\[ 10.7 = 0.5 v_i \left(\frac{45}{v_i \cos 30^{\circ}}\right) – 4.9 \left(\frac{45}{v_i \cos 30^{\circ}}\right)^2 \] Substitute the expression for \( t \) into the vertical equation.
\[ v_i \approx 29.4\,\text{m/s} \] Algebraic manipulation (no calculus) gives the speed; evaluate numerically.
\[ \boxed{v_i \approx 29.4\,\text{m/s}} \] Initial launch speed of the soccer ball.

Part (c): Velocity at wall

Derivation/Formula Reasoning
\[ v_x = v_i \cos 30^{\circ} \] Horizontal component remains constant.
\[ v_x \approx 25.5\,\text{m/s} \] Insert \( v_i = 29.4\,\text{m/s} \).
\[ v_y = v_i \sin 30^{\circ} – g t \] Vertical component at the wall.
\[ v_y \approx -2.62\,\text{m/s} \] Negative sign indicates downward motion.
\[ v = \sqrt{v_x^{2} + v_y^{2}} \] Magnitude of the resultant velocity vector.
\[ v \approx 25.6\,\text{m/s} \] Numerical magnitude.
\[ \phi = \tan^{-1}\!\left(\frac{|v_y|}{v_x}\right) \approx 5.9^{\circ} \] Angle the velocity makes with the horizontal; below because \( v_y < 0 \).
\[ \boxed{v \approx 25.6\,\text{m/s\, at }5.9^{\circ}\text{ below horizontal}} \] Complete velocity description as it clears the wall.

Part (d): 55.5° launch

Derivation/Formula Reasoning
\[ t_2 = \frac{\Delta x}{v_i \cos 55.5^{\circ}} \] Time to reach the wall with the higher angle.
\[ t_2 \approx 2.70\,\text{s} \] Evaluate using \( v_i = 29.4\,\text{m/s} \).
\[ y_2 = y_0 + v_i \sin 55.5^{\circ} t_2 – \tfrac12 g t_2^{2} \] Vertical position at the wall for the new angle.
\[ y_2 \approx 31.3\,\text{m} \] Numerical height above ground.
\[ \boxed{y_2 > 9.0\,\text{m} \;\text{(clears wall)}} \] The ball still passes far above the castle wall.

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!
Quick Actions ?
×

Topics in this question

Join 1-to-1 Elite Tutoring

See how Others Did on this question | Coming Soon

\(1.77\,\text{s}\)
\(29.4\,\text{m/s}\)
\(25.6\,\text{m/s at }5.9^{\circ}\text{ below horizontal}\)
\(31.3\,\text{m}>9.0\,\text{m (clears)}\)

Nerd Notes

Discover the world's best Physics resources

Continue with

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

Error Report

Sign in before submitting feedback.

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

Phy Pro

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

$11.99

per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

Physics is Hard, But It Doesn't Have to Be

We crafted the ultimate A.P Physics 1 course that simplifies everything so you can learn faster and score higher. 

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.

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.

Feeling uneasy about your next physics test? We help boost grades in 3 lesson or less.

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