A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 95 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 max. ht. and acceleration is -9.8 m/s^2.
To determine the speed of the rock at its peak and its acceleration, we can use the principles of projectile motion.
First, let's tackle the speed at the peak of the rock's trajectory. At the peak, the rock momentarily comes to a stop before changing direction and falling back down. Therefore, its speed at the peak is 0 mph (it is momentarily at rest).
Next, let's determine the magnitude and direction of its acceleration. The acceleration of a projectile near the surface of the Earth can be approximated as the acceleration due to gravity, which is approximately 9.8 m/s^2 downward. To convert this to mph^2, we use the following conversion factor:
1 m/s^2 ≈ 2.237 mph^2.
So, the acceleration due to gravity can be approximated as:
9.8 m/s^2 × 2.237 mph^2/m/s^2 ≈ 21.9 mph^2.
Therefore, the magnitude of the rock's acceleration at the peak is approximately 21.9 mph^2, and its direction is downward (negative direction).
To summarize:
- The speed of the rock at the peak is 0 mph (momentarily at rest).
- The magnitude of its acceleration at the peak is approximately 21.9 mph^2, and its direction is downward.