Problem 21 - College Physics

Question

\(\cdot\) The champion jumper of the insect world. 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. \(424,31\) July \(2003, \mathrm{p} .509\) ) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

Step-by-Step Solution

Answer

(a) Takeoff speed: approximately 3.99 m/s. (b) Horizontal distance: approximately 1.46 meters.

Step 1: Use Kinematic Relationship for Maximum Height

For projectile motion, the maximum height (h) can be calculated using the vertical component of the initial velocity (v_y) and the acceleration due to gravity (g). The formula is: \[h = \frac{v_{y}^2}{2g}\]Given that the maximum height h = 58.7 cm = 0.587 m and g = 9.8 m/s², we need to find the vertical component of the velocity \(v_{y}\):\[v_{y}^2 = 2gh = 2 \times 9.8 \times 0.587\]\[v_{y} = \sqrt{2 \times 9.8 \times 0.587} = 3.388\, \text{m/s}\]We have found the vertical component of the initial velocity.

Step 2: Calculate the Total Initial Velocity

The initial velocity, v, of the froghopper's jump can be calculated by considering both the vertical and horizontal components. The vertical component was already calculated as \(v_{y} = 3.388\, \text{m/s}\) and we are given the angle of the jump \(\theta = 58.0^{\circ}\). Use the relation:\[v_y = v \sin(\theta)\]Solving for v:\[v = \frac{v_y}{\sin(\theta)} = \frac{3.388}{\sin(58^{\circ})}\]\[v \approx 3.988\, \text{m/s}\]This is the takeoff speed of the froghopper.

Step 3: Determine the Horizontal Distance Traveled

The horizontal distance \(x\) covered by the froghopper can be calculated using the horizontal component of the initial velocity \(v_x\), the takeoff speed \(v\), and the time of flight. First, calculate \(v_x\): \[v_x = v \cos(\theta) = 3.988 \cos(58^{\circ})\]\[v_x \approx 2.111\, \text{m/s}\]The total time of the flight (T) is given by: \[T = \frac{2v_y}{g} = \frac{2 \times 3.388}{9.8}\]\[T \approx 0.692\, \text{s}\]The horizontal distance can now be calculated as: \[x = v_x \cdot T = 2.111 \times 0.692\]\[x \approx 1.46\, \text{m}\]Thus, the horizontal distance covered by the froghopper is approximately 1.46 meters.

Key Concepts

Kinematic Equations

Kinematic equations play a crucial role in analyzing projectile motion. They describe the motion of objects moving under the influence of gravity, without any other forces such as air resistance affecting the path. The main equations relate variables such as displacement, initial velocity, time, and acceleration. In the case of projectile motion, it's important to separate motion into horizontal and vertical components, because gravity only influences the vertical motion directly. Commonly used equations include:

• \( v = u + at \)
• \( s = ut + \frac{1}{2}at^2 \)
• \( v^2 = u^2 + 2as \)

Here, \(u\) is the initial velocity, \(v\) is the final velocity, \(a\) is the acceleration (\(g\), due to gravity, which is \(-9.8 \text{ m/s}^2\) downward), \(s\) is the displacement, and \(t\) is time. Each equation relies on consistent units, and breaking complex motions into simpler components often simplifies the calculations, as seen with the froghopper's jump.

Initial Velocity Calculation

Calculating the initial velocity of a projectile, like our froghopper's leap, involves understanding both vertical and horizontal components. Given the angle at which the leap occurs, the initial velocity is divided into:

• Vertical component (\(v_y\)): \(v_y = v \sin(\theta)\)
• Horizontal component (\(v_x\)): \(v_x = v \cos(\theta)\)

To find the initial velocity of the froghopper, we used the maximum height to determine the vertical component. For the froghopper, the maximum height formula was used with the value \(h = 0.587 \text{ m}\) to find \(v_y\). Then, using the angle \(58.0^{\circ}\), the overall initial velocity (\(v\)) was calculated by rearranging \(v_y = v \sin(\theta)\) to solve for \(v\), resulting in approximately \(3.988 \text{ m/s}\). This example shows how vector decomposition is vital in problems involving angled projections.

Maximum Height Calculation

The calculation of maximum height in projectile motion requires focusing on the vertical motion only. At maximum height, the vertical velocity becomes zero, which simplifies calculations. The equation \(h = \frac{v_y^2}{2g}\) provides maximum height when knowing the vertical component of initial velocity \(v_y\) and acceleration due to gravity \(g\).In this exercise, the vertical velocity was found using the equation rearranged to \(v_y = \sqrt{2gh}\). Substituting given values \(g = 9.8 \text{ m/s}^2\) and \(h = 0.587 \text{ m}\) yielded \(v_y = 3.388 \text{ m/s}\). This method underscores how peak height assists in understanding initial vertical motion, a key facet in addressing many projectile challenges.

Horizontal Distance Calculation

Determining the horizontal distance in projectile motion combines understanding both horizontal velocity and the total time of flight. Horizontal motion analysis simplifies to one core formula: \(x = v_x \cdot T\), where \(v_x\) is the horizontal velocity component and \(T\) the total time of travel.For the froghopper leap, after finding \(v_x = v \cos(58^{\circ})\), we computed \(v_x\) as approximately \(2.111 \text{ m/s}\). The total time \(T\) is derived from the vertical motion as time to ascend and descend. It was calculated via \(T = \frac{2v_y}{g}\), giving \(0.692 \text{ s}\). Finally, substituting \(v_x\) and \(T\) into \(x = v_x \cdot T\), the result was a horizontal distance of approximately \(1.46 \text{ m}\). This demonstrates how linked vertical and horizontal calculations are essential in understanding overall displacement in projectiles.