A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 91 mph. When the rock reaches its peak, how fast is it going and what is the magnitude and direction of its acceleration? Ignore air drag
when the rock reaches its peak, its velocity is zero, no matter how fast it started out.
The only force acting on it is gravity, so the acceleration is a constant -32ft/s^2 at all times.
To find the speed at the peak and the magnitude and direction of acceleration, we need to analyze the motion of the rock.
Let's break it down into two main phases: upward motion and downward motion.
1. Upward motion:
- At the moment the rock leaves the person's hand, it is going straight up with a velocity of 91 mph (assume mph stands for miles per hour).
- As the rock moves upward, it slows down due to the force of gravity pulling it downward.
- At the peak, the rock momentarily stops completely, briefly achieving zero velocity before reversing direction.
2. Downward motion:
- After reaching the peak, the rock starts moving downward under the influence of gravity.
- It gains speed as it accelerates downward due to the gravitational force.
- The magnitude of acceleration remains constant throughout (approximately 32.2 ft/s^2 or 9.8 m/s^2, considering the acceleration due to gravity near the Earth's surface).
To calculate the speed of the rock at the peak, we know that the magnitude of its velocity decreases by 32.2 ft/s or 9.8 m/s every second. Since speed is a scalar quantity, it will have the same numerical value as velocity.
Knowing this, we can calculate the time it takes for the rock to reach its peak by dividing the initial velocity of 91 mph (or converted to ft/s or m/s) by the acceleration due to gravity (-32.2 ft/s^2 or -9.8 m/s^2). The negative sign indicates the opposite direction of the gravitational force.
Let's convert the initial velocity from 91 mph to ft/s or m/s and calculate the time to reach the peak:
91 mph = 40.56 m/s or 133.14 ft/s
Dividing the initial velocity by the acceleration due to gravity, we get:
Time to reach the peak = (-40.56 m/s) / (-9.8 m/s^2) ≈ 4.14 s
Now that we know the time, we can find the speed at the peak by subtracting the product of time and acceleration due to gravity from the initial velocity:
Speed at the peak = 40.56 m/s - 9.8 m/s^2 * 4.14 s ≈ 1.21 m/s or 4.00 ft/s
Therefore, the speed of the rock at the peak is approximately 1.21 m/s or 4.00 ft/s.
Regarding the acceleration at the peak, it is zero since the rock momentarily comes to rest before reversing its direction. The magnitude of acceleration throughout the motion remains constant at approximately 32.2 ft/s^2 or 9.8 m/s^2, directed downwards.
To summarize:
- Speed at the peak: approximately 1.21 m/s or 4.00 ft/s
- Magnitude of acceleration: approximately 32.2 ft/s^2 or 9.8 m/s^2
- Direction of acceleration: downward