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?
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Projectile motion range is derived as follows:
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}.
3 clay balls, labeled A, B, and C are launched from the same height at the same speed as shown below. A is launched at 30° above horizontal, B is launched horizontally, and C is launched 30° below the horizontal. They all hit the wall (before reaching the ground) in times tA, tB, and tC respectively. Rank these times from least to greatest.
Three identical rocks are launched with identical speeds from the top of a platform of height h0.
Which of the following correctly relates the magnitude vy of the vertical component of the velocity of each rock immediately before it hits the ground?
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The acceleration of the javelin during the throw.
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A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.
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The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
A person shoots a basket ball with a speed of 12 m/s at an angle of 35° above the horizontal. If the person is 2.4 m tall and the hoop is 3.05 m above the ground, how far back must the person stand in order to make the shot?
You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.
You must split an apple resting on top of you friend’s head from a distance of 27 m. When you aim directly at the apple, the arrow is horizontal. At what angle should you aim the arrow to hit the apple if the arrow travels at a speed of 35 m/s?
A diver springs upward from a diving board. At the instant she contacts the water her speed is 8.90 m/s, and her body is extended at an angle of 75.0° with respect to the horizontal surface of the water. At this instant her vertical displacement is -3.00 m, where downward is the negative direction. Determine her initial velocity, both magnitude and direction.
Which of the following statements about the acceleration due to gravity is TRUE?
A ball is kicked at a speed of vo an angle ø above the horizontal. The ball travels 25 meters horizontally. If the ball is kicked at 2vo, what will the horizontal displacement be?
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Home » 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?
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Kinematics | Forces |
---|---|
\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |
v = v_i + a \cdot t | F_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 Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum m \cdot r^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k \cdot 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} |
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 (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.
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 |