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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

Two piers, \(A\) and \(B,\) are located on a river; \(B\) is \(1500 \mathrm{~m}\) down- stream from \(A\) (Fig. E3.38). Two friends must make round trips from pier \(A\) to pier \(B\) and return. One rows a boat at a constant speed of \(4.00 \mathrm{~km} / \mathrm{h}\) relative to the water; the other walks on the shore at a constant speed of \(4.00 \mathrm{~km} / \mathrm{h}\). The velocity of the river is \(2.80 \mathrm{~km} / \mathrm{h}\) in the direction from \(A\) to \(B\). How much time does it take each person to make the round trip?

A man stands on the roof of a 15.0 -m-tall building and throws a rock with a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

In Canadian football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is \(10.0 \mathrm{ft}\) above the ground, and the ball is kicked from ground level, \(36.0 \mathrm{ft}\) horizontally from the bar (Fig. \(\mathbf{P 3 . 6 0}\) ). Football regulations are stated in English units, but convert them to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it is just to clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\).

A "moving sidewalk" in an airport terminal moves at \(1.0 \mathrm{~m} / \mathrm{s}\) and is \(35.0 \mathrm{~m}\) long. If a woman steps on at one end and walks at \(1.5 \mathrm{~m} / \mathrm{s}\) relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

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?
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