- The ball reaches the ground in approximately \(\boxed{3.03 \text{ s}}\) at a speed of \(\boxed{29.70 \text{ m/s}}\).
- The hammer and feather land simultaneously in a vacuum because acceleration is independent of mass.
- The ball's acceleration is constant at \(-9.8 \text{ m/s}^2\) (downward) throughout its flight, even when velocity is zero at the peak.
- The skydiver's average acceleration decreases over time due to increasing air resistance, eventually reaching \(\boxed{0 \text{ m/s}^2}\) at terminal velocity (54 m/s).
Explanation
- Problem 1: Free Fall from a Tower
For all parts of this problem, we define the upward direction as positive (+). Gravity, acting downward, is therefore negative (\(-9.8 \text{ m/s}^2\)).
- Dropped Ball Analysis:
AA ball is dropped from a height \(h = 45 \text{ m}\) with initial velocity \(u = 0\). Using the kinematic equation \(h = ut + \frac{1}{2}gt^2\), where \(g = 9.8 \text{ m/s}^2\) (taking the magnitude for the fall distance), we have:
\[45 = 0 + \frac{1}{2}(9.8)t^2\]
\[45 = 4.9t^2\]
\[t^2 = \frac{45}{4.9} \approx 9.1837\]
\[t \approx 3.03 \text{ s}\]
The time taken to reach the ground is approximately \(3.03 \text{ s}\).
BUsing \(v = u + gt\), the speed at impact is:
\[v = 0 + 9.8 \times 3.0305 \approx 29.70 \text{ m/s}\]
The speed at impact is approximately \(29.70 \text{ m/s}\).
- Problem 2: Hammer and Feather on the Moon
- Moon Experiment:
On the Moon, where there is no atmosphere, a hammer and a feather land at exactly the same time. According to Newton's Second Law, the net force on an object in free fall is \(F_{net} = mg\). The acceleration \(a\) is determined by \(a = \frac{F_{net}}{m} = \frac{mg}{m} = g\). Because the acceleration due to gravity (\(g\)) is the same for all objects regardless of their mass, they will accelerate at the same rate and hit the ground simultaneously.
- Problem 3: Ball Thrown Upward
- Ball Thrown Upward:
When a ball is thrown straight up (with upward defined as positive), its velocity changes from positive to negative, but the acceleration remains constant at \(-9.8 \text{ m/s}^2\) (downward) throughout the flight.
| Moment | Velocity | Acceleration |
| :--- | :--- | :--- |
| Rising, halfway to the top | Positive (+) | \(-9.8 \text{ m/s}^2\) (downward) |
| At the very top | \(0\) | \(-9.8 \text{ m/s}^2\) (downward) |
| Falling, halfway down | Negative (-) | \(-9.8 \text{ m/s}^2\) (downward) |
- Problem 4: Skydiver Motion Analysis
- Skydiver Analysis:
AAverage acceleration (\(a_{avg}\)) is the change in velocity over time: \(a_{avg} = \frac{\Delta v}{\Delta t}\).
- From \(t=0\) to \(t=2 \text{ s}\): \(a_{avg} = \frac{15 - 0}{2 - 0} = 7.5 \text{ m/s}^2\).
- From \(t=10\) to \(t=12 \text{ s}\): \(a_{avg} = \frac{54 - 50}{12 - 10} = 2.0 \text{ m/s}^2\).
BThese values differ because, as the skydiver's speed increases, air resistance (drag) also increases. Initially, drag is negligible and acceleration is close to \(g\). As speed rises, the drag force grows, opposing gravity and reducing the net force (\(F_{net} = F_{gravity} - F_{drag}\)), which in turn reduces the acceleration.
CAt \(t=14 \text{ s}\), the velocity is \(54 \text{ m/s}\), the same as it was at \(t=12 \text{ s}\). Since the velocity is constant, the acceleration at \(t=14 \text{ s}\) is \(0 \text{ m/s}^2\). The skydiver has reached terminal velocity, where the drag force exactly balances the weight of the skydiver.
Examples
Understanding kinematics is crucial in fields like aerospace engineering and sports science. For instance, designing a safe landing system for a drone involves calculating the exact time and speed at which it will touch down based on its altitude, much like the dropped ball problem. Similarly, elite athletes analyze the velocity and acceleration of their movements to maximize jump height or sprinting speed, illustrating how these physics principles translate directly into real-world performance.