Cuban tree frogs are skilled jumpers. They can launch themselves at 4.5 meters per second at an angle of 25° above horizontal. Assuming the frog is jumping on level ground, how far will the frog travel in a single jump, what is the frog’s hang time in the air during a single jump, and what is the maximum height above the ground that the frog will achieve?
The Correct Answer and Explanation is:
To solve for the distance, hang time, and maximum height of the Cuban tree frog’s jump, we need to break the problem into horizontal and vertical components, considering projectile motion. Here’s how to approach it:
Given:
- Initial velocity (( v_0 )) = 4.5 m/s
- Launch angle (( \theta )) = 25°
- Acceleration due to gravity (( g )) = 9.8 m/s²
Step 1: Resolve the initial velocity into horizontal and vertical components.
- Horizontal velocity:
[ v_{0x} = v_0 \cos(\theta) ]
[ v_{0x} = 4.5 \times \cos(25^\circ) \approx 4.5 \times 0.9063 \approx 4.08 \, \text{m/s} ] - Vertical velocity:
[ v_{0y} = v_0 \sin(\theta) ]
[ v_{0y} = 4.5 \times \sin(25^\circ) \approx 4.5 \times 0.4226 \approx 1.90 \, \text{m/s} ]
Step 2: Hang time (time the frog stays in the air).
The time to reach the maximum height and fall back down is determined by the vertical motion, where the time to reach maximum height is when the vertical velocity becomes zero.
- Time to reach the peak height:
[ t_{\text{up}} = \frac{v_{0y}}{g} ]
[ t_{\text{up}} = \frac{1.90}{9.8} \approx 0.194 \, \text{seconds} ]
Since the time to go up is equal to the time to come down, the total hang time is:
[ t_{\text{total}} = 2 \times t_{\text{up}} \approx 2 \times 0.194 \approx 0.388 \, \text{seconds} ]
Step 3: Maximum height above the ground.
The maximum height is determined by the vertical motion, using the formula for the displacement in the y-direction:
[ h_{\text{max}} = \frac{v_{0y}^2}{2g} ]
[ h_{\text{max}} = \frac{1.90^2}{2 \times 9.8} \approx \frac{3.61}{19.6} \approx 0.18 \, \text{meters} ]
Step 4: Range (distance traveled horizontally).
The horizontal distance (range) is calculated using the formula:
[ \text{Range} = v_{0x} \times t_{\text{total}} ]
[ \text{Range} = 4.08 \times 0.388 \approx 1.59 \, \text{meters} ]
Summary of Results:
- The frog will travel approximately 1.59 meters.
- The hang time is approximately 0.388 seconds.
- The maximum height the frog will achieve is approximately 0.18 meters above the ground.
These calculations show how the frog’s jumping capabilities depend on its initial velocity, launch angle, and gravity. The horizontal distance covered is a product of the frog’s horizontal speed and the total time it is in the air. The vertical motion determines the hang time and maximum height.