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Chapter 3: Problem 14

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of \(58.7 \mathrm{~cm}\) above the level ground. (See Nature, Vol. 424, July \(31,2003,\) p. 509.) Neglect air resistance in answering the following. (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world- record leap?

Short Answer

Expert verified
The initial velocity or take-off speed for the leap is approximately \(2.94 m/s\). The horizontal distance the froghopper covered for this world-record leap is approximately \(1.04 m\).

Step by step solution

01

Calculate Initial Velocity

Use the equation of the maximum height of a projectile, where \(H\) is the maximum height, \(v\) is the initial velocity, and \(\theta\) is the angle of projection. This equation is given by: \(H = \frac{v^2 \cdot \sin^2(\theta)}{2g}\) where \(g\) is the acceleration due to gravity which is approximately \(9.8 m/s^2\). From the given problem, \(\theta = 58.0^{\circ}\) and \(H = 58.7 cm = 0.587 m\). Substitute \(H\), \(\theta\) and \(g\) in the aforementioned equation and solve for \(v = \sqrt{\frac{2gH}{\sin^2(\theta)}}\).
02

Calculate Horizontal Distance

Now, calculate the time taken to reach maximum height using the formula \(t = \frac{v \cdot \sin(\theta)}{g}\). Based on the feature of symmetrical trajectory in projectile motion, the total time of flight (time to go up and come down) will be \(2t\). Then, calculate the horizontal distance using the formula \(d = v \cdot \cos(\theta) \cdot t_{total}\).

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Key Concepts

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

Initial Velocity Calculation
Understanding the initial velocity of a projectile is crucial in analyzing its motion. To find this, you need to know the maximum height reached and the angle of projection. Using the equation
\( H = \frac{v^2 \cdot \sin^2(\theta)}{2g} \),
where \( H \) is the maximum height, \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity, we can solve for \( v \). This equation rearranges to
\( v = \sqrt{\frac{2gH}{\sin^2(\theta)}} \).
You substitute the known values for \( H \), \( \theta \), and \( g \) to calculate \( v \). For example, if a froghopper jumps at a \(58^\circ\) angle and reaches 58.7 cm, we convert centimeters to meters and plug in \(9.8 m/s^2\) for gravity to get the initial velocity.
Maximum Height of a Projectile
The maximum height a projectile can reach is a function of its initial velocity and the angle at which it is projected. The formula
\( H = \frac{v^2 \cdot \sin^2(\theta)}{2g} \)
shows that height is proportional to the square of the initial velocity and the square of the sine of the projection angle. Gravity inversely affects this height. It is imperative to note that only the vertical component of the initial velocity (that's the \( v \sin(\theta) \) part) influences the maximum height. For instance, in the case of the leaping froghopper, the seemingly small initial velocity, when properly directed at a large angle, can still produce impressive heights.
Horizontal Distance in Projectile Motion
Projectile motion consists of two components: vertical and horizontal. While the vertical component determines the maximum height, the horizontal component dictates how far the projectile will travel. To calculate the horizontal distance, you first find the time taken to reach the maximum height with the formula
\( t = \frac{v \cdot \sin(\theta)}{g} \),
and then double it to get the total time of flight鈥� since the ascent and descent times are equal. Next, use the total time of flight and the horizontal velocity, which is the initial velocity times the cosine of the projection angle, \( v \cos(\theta) \), to calculate the horizontal distance with
\( d = v \cdot \cos(\theta) \cdot t_{total} \).
This horizontal distance is a critical factor in many applications, such as sports and engineering, where the projections must be precise and optimal.

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Most popular questions from this chapter

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. \(\mathbf{P 3 . 6 3 )}\). The takeoff ramp was inclined at \(53.0^{\circ},\) the river was \(40.0 \mathrm{~m}\) wide, and the far bank was \(15.0 \mathrm{~m}\) lower than the top of the ramp. The river itself was \(100 \mathrm{~m}\) below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path \(90.0 \mathrm{~m}\) above the ground and has a speed of \(64.0 \mathrm{~m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of \(2.1 \mathrm{~m}\) from this point (Fig. E3.19). If you toss the coin with a velocity of \(6.4 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal, the coin will land in the dish. Ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

The coordinates of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{~m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{~m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{~m} / \mathrm{s}^{2} .\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{~s}\). (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=2.0 \mathrm{~s}\). (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{~s}\). At this instant, is the bird's speed increasing, decreasing, or not changing? Is the bird turning? If so, in what direction?

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?
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