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It takes 6seconds for a stone dropped from rest to hit the bottom of a well what is its final speeds?
Using the equation of motion s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is acceleration, and t is time, we can find the final speed. We know that initial velocity, u, is 0 as the stone is dropped from rest. The acceleration, a, would be due to gravity (-9.8 m/s^2). Rearrange the equation to find final speed, v: v = at. Plug in the values: v = 9.8 m/s^2 * 6s = 58.8 m/s. So, the final speed of the stone when it hits the bottom of the well is 58.8 m/s.
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What is the velocity of a dropped object after it has fallen for 3.0?
The final velocity of a dropped object can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time the object has fallen. Plugging in the values, the final velocity of a dropped object after falling for 3.0 seconds would be 29.43 m/s.
An object is dropped from rest at a height of 128 m. Find the distance it falls during its final second in the air.?
In the final second, the object accelerates at a constant rate of 9.8 m/s^2. Using the equation of motion, the distance fallen during the final second can be calculated as 1/2 * 9.8 * 1^2 = 4.9 meters.
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The final velocity of the ball when it hits the ground can be calculated using the equation: final velocity = initial velocity + (acceleration due to gravity * time). Assuming the ball was dropped from rest, the initial velocity would be 0 m/s. With the acceleration due to gravity being approximately 9.8 m/s^2, the final velocity would be 32.34 m/s.
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It takes 6 seconds for a stone dropped from rest to hit the bottom of a well what is its final speed?
60 m/s
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What is the velocity of a dropped object after it has fallen for 3.0?
The final velocity of a dropped object can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time the object has fallen. Plugging in the values, the final velocity of a dropped object after falling for 3.0 seconds would be 29.43 m/s.
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An object is dropped from rest at a height of 128 m. Find the distance it falls during its final second in the air.?
In the final second, the object accelerates at a constant rate of 9.8 m/s^2. Using the equation of motion, the distance fallen during the final second can be calculated as 1/2 * 9.8 * 1^2 = 4.9 meters.
