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Chapter 3: Problem 61

(a) Prove that a projectile launched at angle \(\alpha_{0}\) has the same horizontal range as one launched with the same speed at angle \(\left(90^{\circ}-\alpha_{0}\right) \cdot(\text { b) A frog jumps at a speed of } 2.2 \mathrm{m} / \mathrm{s} \text { and lands } 25 \mathrm{cm}\) from its starting point. At which angles above the horizontal could it have jumped?

Short Answer

Expert verified
15.4掳 or 74.6掳.

Step by step solution

01

Understand the Problem

We need to prove the horizontal range for a projectile launched at an angle \(\alpha_0\) is the same as for one launched at \(90^\circ - \alpha_0\). Then find possible angles for a frog's jump given its speed and horizontal distance.
02

Horizontal Range Formula

The formula for the horizontal range \(R\) of a projectile is given by \(R = \frac{v_0^2 \sin(2\alpha)}{g}\), where \(v_0\) is the launch speed, \(\alpha\) is the launch angle, and \(g\) is the acceleration due to gravity.
03

Prove Range Equivalence

To prove equivalence, compute the range for \(\alpha_0\) and \(90^\circ - \alpha_0\). Substitute \(2\alpha_0\) in the range formula: \(\sin(2\alpha_0) = \sin(2(90^\circ - \alpha_0)) = \sin(180^\circ - 2\alpha_0) = \sin(2\alpha_0)\). Thus, \(R(\alpha_0) = R(90^\circ - \alpha_0)\).
04

Calculate Launch Speed and Distance for Frog

Given the frog jumps at \(v_0 = 2.2 \text{ m/s}\) and covers a horizontal distance \(R = 0.25 \text{ m}\). Use the range equation \(R = \frac{v_0^2 \sin(2\alpha)}{g}\). Rearrange to find \(\alpha\): \(\sin(2\alpha) = \frac{R \cdot g}{v_0^2}\).
05

Solve for Possible Angles

Substitute the known values: \(\sin(2\alpha) = \frac{0.25 \times 9.81}{2.2^2} \approx 0.512 \). Solve \(2\alpha = \sin^{-1}(0.512)\) giving \(2\alpha \approx 30.8^\circ\), hence \(\alpha \approx 15.4^\circ\). Also, use \(180^\circ - 2\alpha \) for the second angle, \(180^\circ - 30.8^\circ = 149.2^\circ\), so \(\alpha \approx 74.6^\circ\).
06

Final Result

The frog could have jumped at angles approximately \(15.4^\circ\) or \(74.6^\circ\) to land 25 cm from its starting point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Range
The horizontal range of a projectile is the distance it travels along the horizontal plane from the launch point to the landing point. This measurement is crucial in understanding how far a projectile can reach given a specific set of initial conditions.

To calculate the horizontal range, you use the formula:
  • \( R = \frac{v_0^2 \sin(2\alpha)}{g} \)
where:
  • \( R \) is the horizontal range,
  • \( v_0 \) is the initial velocity,
  • \( \alpha \) is the launch angle,
  • \( g \) is the acceleration due to gravity, typically \(9.81\, \text{m/s}^2\).
Understanding how this formula works is crucial when dealing with various physics problems related to projectile motion. The range depends significantly on both the speed and angle of launch, making these parameters vital in the planning stages of anything from sports to space missions.
Launch Angle
The launch angle of a projectile is the initial angle at which it is projected relative to the horizontal axis. This angle plays a crucial role in determining the trajectory and range of the projectile.

A projectile's launch angle can dramatically affect how far it travels:
  • When the launch angle is \(45^\circ\), the projectile typically achieves the maximum possible range when air resistance is negligible.
  • Angles less than \(45^\circ\) result in a shorter range but a flatter trajectory.
  • Angles greater than \(45^\circ\) create a steeper trajectory, leading to a higher peak but a shorter range.
Understanding the optimal launch angle is essential for maximizing the range while taking various practical factors into account, such as initial speed and environmental conditions.
Kinematic Equations
Kinematic equations are mathematical formulas that describe the motion of objects without considering the causes of this motion (such as forces). In projectile motion, these equations help us determine a projectile's future position and velocity at any given time.

For a projectile launched with an initial speed \( v_0 \) at an angle \( \alpha \), we consider:
  • Horizontal motion: \( x = v_0 \cdot \cos(\alpha) \cdot t \)
  • Vertical motion: \( y = v_0 \cdot \sin(\alpha) \cdot t - \frac{1}{2}gt^2 \)
where:
  • \( x \) and \( y \) are the horizontal and vertical positions,
  • \( t \) is the time elapsed,
  • \( g \) is the gravitational acceleration.
Using these formulas effectively allows us to predict projectile paths, facilitating everything from ballistics to sports tactics.
Projectile Trajectory
The trajectory of a projectile is the path it follows after being launched. This path is typically a curve or arc known as a parabola, due to the constant acceleration downwards caused by gravity.

A projectile's trajectory is influenced by the initial conditions of the launch, specifically the:
  • Launch speed (\( v_0 \))
  • Launch angle (\( \alpha \))
  • Height from which it is launched
The motion can be broken into two components:
  • Horizontal motion, which is uniform (constant velocity)
  • Vertical motion, which is uniformly accelerated (due to gravity)
Understanding projectile trajectory is essential in various fields like engineering, sports, and gaming, where predicting and controlling the flight path is critical.

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Most popular questions from this chapter

A jet plane is flying at a constant altitude. At time \(t_{1}=0\) it has components of velocity \(v_{x}=90 \mathrm{m} / \mathrm{s}, v_{y}=110 \mathrm{m} / \mathrm{s}\) . At time \(t_{2}=30.0 \mathrm{s}\) the components are \(v_{x}=-170 \mathrm{m} / \mathrm{s}, v_{y}=40 \mathrm{m} / \mathrm{s}\) . (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2}\) . How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

A pistol that fires a signal flare gives it an initial velocity (muzzle velocity) of 125 \(\mathrm{m} / \mathrm{s}\) at an angle of \(55.0^{\circ}\) above the borizontal. You can ignore air resistance. Find the flare's maximum height and the distance from its firing point to its landing point if it is fired (a) on the level salt flats of Utah, and (b) over the flat Sea of Tranquility on the Moon, where \(g=1.67 \mathrm{m} / \mathrm{s}^{2}\) .

Two students are canoeing on a river. While heading effort? upstream, they accidentally drop an emply bottle oveabound. They then continue paddling for 60 minutes, reaching a point 20 \(\mathrm{km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they tum around and head downstram. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 \(\mathrm{km}\) down- stream from the tum-around point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 90.0 \(\mathrm{km} / \mathrm{h}\) , to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.
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