A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 81 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. Express your answer using appropriate mks units
at the peak, it has stopped, so speed=0
acceleration is just -g, as usual.
To find the speed of the rock when it reaches its peak, we need to analyze the motion of the rock as it moves up and then falls back down.
Step 1: Determine the vertical speed at the peak point:
When the rock reaches its peak, its vertical velocity becomes zero. This is because at the highest point of its motion, the rock momentarily stops before it starts falling back down. Therefore, the speed of the rock at the peak is 0 mph.
Step 2: Find the acceleration of the rock:
The acceleration of an object under free fall near the Earth's surface is approximately equal to the acceleration due to gravity, which is 9.8 m/s^2. This acceleration acts downward, towards the center of the Earth.
Step 3: Determine the magnitude and direction of the acceleration:
The magnitude of the acceleration is given by the absolute value of 9.8 m/s^2, which is 9.8 m/s^2. The direction of the acceleration is negative because it acts downward. Therefore, the magnitude of the acceleration is 9.8 m/s^2, and its direction is downward.
In summary, the rock's speed when it reaches its peak is 0 mph, and the magnitude and direction of its acceleration are 9.8 m/s^2 downward.
To determine the velocity of the rock when it reaches its peak, we need to consider its motion under the influence of gravity. When the rock is at its peak, it momentarily comes to a stop before starting to fall back down. At this point, the velocity of the rock is zero.
Gravity acts as an acceleration on the rock, pulling it downwards. The magnitude of this acceleration is constant near the Earth's surface and is equal to approximately 9.8 meters per second squared (m/s^2) or 32.2 feet per second squared (ft/s^2).
Since the rock is initially thrown straight up, the acceleration due to gravity acts in the opposite direction to the motion of the rock. Therefore, its direction is considered negative.
To find the magnitude and direction of the rock's acceleration when it reaches its peak, we can use the value of 9.8 m/s^2 or 32.2 ft/s^2.
Next, let's convert the initial velocity of the rock from mph to m/s. To convert from mph to m/s, we can use the conversion factor 1 mph = 0.44704 m/s.
Given that the initial velocity of the rock is 81 mph, we can multiply it by the conversion factor:
81 mph * 0.44704 m/s = 36.07124 m/s
Therefore, the initial velocity of the rock is approximately 36.07 m/s.
At the peak of its trajectory, the velocity of the rock is zero, indicating that its speed has decreased to zero.
The acceleration due to gravity remains the same throughout the motion, so its magnitude at the peak is still 9.8 m/s^2. However, the acceleration now acts in the positive direction since the rock has reversed its direction of motion.
Therefore, the velocity of the rock at its peak is 0 m/s, and the magnitude of its acceleration is 9.8 m/s^2 in the positive direction.