A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling? A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. What is the pelican’s speed
The Correct Answer and Explanation is:
To solve these two projectile motion problems, we use principles of kinematics. In both scenarios, the horizontal velocity remains constant, and the vertical motion is influenced by gravity. Here’s how to calculate the initial horizontal velocity in each case.
Problem 1: Baseball off a desk
Given:
- Height of desk (vertical displacement), h=0.70 mh = 0.70 \, \text{m}h=0.70m
- Horizontal distance to the floor (range), R=0.25 mR = 0.25 \, \text{m}R=0.25m
- Acceleration due to gravity, g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
Step 1: Find the time to fall
Use the formula:y=12gt2y = \frac{1}{2} g t^2y=21gt2
Solving for ttt:t=2yg=2×0.709.8≈0.143≈0.378 st = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 0.70}{9.8}} \approx \sqrt{0.143} \approx 0.378 \, \text{s}t=g2y=9.82×0.70≈0.143≈0.378s
Step 2: Calculate horizontal velocityvx=horizontal distancet=0.250.378≈0.661 m/sv_x = \frac{\text{horizontal distance}}{t} = \frac{0.25}{0.378} \approx 0.661 \, \text{m/s}vx=thorizontal distance=0.3780.25≈0.661m/s
Answer:
The baseball was rolling at approximately 0.66 m/s.
Problem 2: Fish dropped by a pelican
Given:
- Height, h=5.4 mh = 5.4 \, \text{m}h=5.4m
- Horizontal distance, R=8.0 mR = 8.0 \, \text{m}R=8.0m
- g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
Step 1: Find time to fallt=2yg=2×5.49.8=1.102≈1.05 st = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 5.4}{9.8}} = \sqrt{1.102} \approx 1.05 \, \text{s}t=g2y=9.82×5.4=1.102≈1.05s
Step 2: Find horizontal speedvx=8.01.05≈7.62 m/sv_x = \frac{8.0}{1.05} \approx 7.62 \, \text{m/s}vx=1.058.0≈7.62m/s
Answer:
The pelican’s speed is approximately 7.6 m/s.
Explanation
In both problems, an object is in projectile motion, meaning it moves under the influence of gravity after initial horizontal movement. A key feature of projectile motion is that horizontal and vertical motions are independent of each other. Gravity affects only the vertical component while the horizontal velocity remains constant in the absence of air resistance.
To solve these problems, we start by calculating how long the object takes to fall from its initial height to the ground. This is purely a vertical motion problem. The time of fall depends only on the height and gravity. We use the kinematic equation y=12gt2y = \frac{1}{2} g t^2y=21gt2, where yyy is the height and ggg is the acceleration due to gravity. Solving for ttt gives us the duration of the fall.
Once we know the time it takes to fall, we turn to the horizontal motion. Since there is no horizontal acceleration, we use the simple relationship v=dtv = \frac{d}{t}v=td, where ddd is the horizontal distance traveled and ttt is the time already found. This gives us the initial horizontal velocity of the object — either how fast the ball was rolling or how fast the pelican was flying.
This process highlights how motion in two dimensions can be broken down into separate, simpler one-dimensional motions. These types of calculations are foundational in physics and help explain a variety of real-world motions from sports to wildlife behavior.
