The minimum velocity is the initial vertical velocity required to reach 1500 m under the influence of gravity.
Step | Formula Derivation | Reasoning |
---|---|---|
1.1 | v_{\text{min}} = \sqrt{2gh} | The minimum velocity required, derived from the equation v^2 = v_0^2 + 2gh, where v_0 is initial velocity and v is final velocity (0 at the top). |
This is the time taken for the rocket to reach its highest point, where its vertical velocity becomes zero.
Step | Formula Derivation | Reasoning |
---|---|---|
2.1 | t = \frac{v_0}{g} | Time to reach maximum height, where v_0 is initial vertical velocity and g is acceleration due to gravity. |
The highest point the rocket reaches when launched vertically at 180 m/s.
Step | Formula Derivation | Reasoning |
---|---|---|
3.1 | h = \frac{v_0^2}{2g} | Maximum height reached, derived from v^2 = v_0^2 – 2gh, with v = 0 at the maximum height. |
Assuming the rocket reaches the spacecraft, we calculate the time and velocity of impact.
Step | Formula Derivation | Reasoning |
---|---|---|
4.1 | v = \sqrt{v_0^2 – 2gh} | Velocity at impact, derived from v^2 = v_0^2 – 2gh. |
4.2 | t = \frac{v_0 – v}{g} | Time of impact, calculated from v = v_0 – gt. |
We will now calculate these values.
Here are the results:
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A car accelerates from rest with an acceleration of 3.5 m/s2 for 10 seconds. After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp 50 m ahead and takes 0.6 seconds to react. After reacting, the driver hits the brakes which slow the car with an acceleration of 7.2 m/s2. Unfortunately, the driver does not stop in time and goes off the 3m high ramp that is angled at 27°.
An object is projected vertically upward from ground level. It rises to a maximum height H . If air resistance is negligible, which of the following must be true for the object when it is at a height H/2 ?
Police officers have measured the length of a car’s tire skid marks to be 23 meters. This particular car is known to decelerate at a constant 7.5 m/s2. What was the car’s initial velocity?
A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?
You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_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 Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients 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. |
k | Spring 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 |
Variable | SI 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)} |
Variable | Derived 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.
Start with the given measurement: \text{5 km}
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}}
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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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