91Ó°ÊÓ

The takeoff speed is crucial in determining how far and high a projectile, like the froghopper, will go. To calculate the takeoff speed, we first need to find the initial vertical velocity component. Here, we use the principle that kinetic energy at the start turns fully into potential energy at the max height. We know the formula:\[ v_y^2 = 2gh \]where

• \( v_y \) is the vertical component of the initial velocity,
• \( g \) is the acceleration due to gravity \( (9.8\, \mathrm{m/s^2}) \),
• \( h \) is the maximum height converted to meters (0.587 m).

By rearranging and plugging in the known values, we find:\[ v_y = \sqrt{2 \times 9.8\, \mathrm{m/s^2} \times 0.587\, \mathrm{m}} \approx 3.385\, \mathrm{m/s} \]Next, use the vertical velocity component to find the takeoff speed \( v_0 \). Since\[ v_y = v_0 \cdot \sin(58.0^\circ) \]we can derive:\[ v_0 = \frac{v_y}{\sin(58.0^\circ)} \approx \frac{3.385\, \mathrm{m/s}}{\sin(58.0^\circ)} \approx 3.98\, \mathrm{m/s} \]This tells us that the froghopper must have had a takeoff speed of about 3.98 m/s.

Vertical motion analysis in projectile motion helps us determine key aspects like maximum height and time of flight. When the froghopper jumps, gravity affects its vertical travel. At the peak of its jump, its vertical velocity is zero. We know the max height is when upward velocity becomes zero. We calculate the time to reach that height using:\[ t_{\text{up}} = \frac{v_y}{g} \approx \frac{3.385\, \mathrm{m/s}}{9.8\, \mathrm{m/s^2}} \approx 0.345\, \mathrm{s} \]This is the time it took to ascend. For symmetric trajectory, total flight time (up and down) is:\[ t = 2 \times t_{\text{up}} \approx 0.69\, \mathrm{s} \]These calculations show how long the froghopper remains airborne, crucial for further determining how far it can jump horizontally. Vertical motion calculations form the backbone of understanding overall projectile dynamics.

Understanding horizontal motion is key to determining the horizontal distance a projectile covers. In our froghopper's leap, we analyze horizontal distance using the constant horizontal velocity and time in motion. The horizontal component of initial velocity is given by:\[ v_{0x} = v_0 \cdot \cos(58.0^\circ) \approx 3.98\, \mathrm{m/s} \cdot \cos(58.0^\circ) \]Calculate:\[ v_{0x} \approx 3.98\, \mathrm{m/s} \cdot 0.5299 \approx 2.11\, \mathrm{m/s} \]Then, knowing total flight time \( t \approx 0.69 \) s, we calculate horizontal distance \( d \) using:\[ d = v_{0x} \cdot t \approx 2.11\, \mathrm{m/s} \times 0.69\, \mathrm{s} \approx 1.44\, \mathrm{m} \]This distance of approximately 1.44 meters represents the froghopper's leap length. Horizontal motion uses constant velocity equations since external forces like drag are typically negligible over such small scales.