A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 76 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.
The velocity is zero at its' peak.
The acceleration = -9.8 m/s^2.
To find the speed and acceleration of the rock when it reaches its peak, we need to analyze the motion of the rock.
First, let's understand the concept of acceleration. Acceleration is the rate at which an object's velocity changes. It is a vector quantity, meaning it has both magnitude and direction.
The magnitude of the acceleration due to gravity on Earth is approximately 9.8 meters per second squared (m/s²) or 32 feet per second squared (ft/s²), directed downward.
Now, let's break down the motion of the rock. When the person throws the rock straight up, it initially has an upward velocity of 76 mph. However, as the rock moves upward, it experiences the acceleration due to gravity, causing its velocity to decrease.
To find the speed at the peak, we need to determine the point where the rock momentarily stops moving upward before starting to fall back down. At this point, the rock's velocity is zero.
To find the time it takes for the rock to reach its peak, we can use the fact that the vertical velocity of the rock decreases at a rate of 9.8 m/s² or 32 ft/s². We can use the following kinematic equation:
v = u + at
where:
v is the final velocity (zero in this case),
u is the initial velocity (76 mph),
a is the acceleration due to gravity (-9.8 m/s² or -32 ft/s²), and
t is the time taken.
Converting the initial velocity to meters per second (m/s), we have:
u = 76 mph = 34 m/s (approximately)
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values, we have:
t = (0 - 34) / -9.8 ≈ 3.47 seconds
Therefore, it takes approximately 3.47 seconds for the rock to reach its peak.
Now, let's find the speed of the rock at its peak. Since the rock stops momentarily at the peak, its speed is zero.
For the acceleration at the peak, we know that the acceleration due to gravity is constant and always directed downward. Therefore, the magnitude of the acceleration at the peak is 9.8 m/s² or 32 ft/s², and its direction is downward.
In summary:
- The speed of the rock at its peak is zero.
- The magnitude of the acceleration at the peak is 9.8 m/s² or 32 ft/s², directed downward.