A ball rolls along a desktop with an uniform velocity of 3 m/s. What is the displacement after 10 seconds have passed?
A ball is thrown upward with an initial velocity of 18 m/s? What is the maximum height the ball reaches if the ball is assumed to leave from the ground?
A ball is dropped from a 35 m tall building. The ball takes how long to impact the ground?
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first part the answer is 30 m. d= rt so distance = rate (3) times the time (10) so the distance is 30
To find the displacement of the ball rolling along the desktop, we would use the formula:
displacement = velocity × time
Given that the velocity of the ball is 3 m/s and the time is 10 seconds, the displacement can be calculated as:
displacement = 3 m/s × 10 s = 30 meters
Therefore, the displacement after 10 seconds is 30 meters.
To find the maximum height reached by a ball thrown upward, we can use the formula:
maximum height = (initial velocity)^2 / (2 × acceleration due to gravity)
Given that the initial velocity of the ball is 18 m/s and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the maximum height as:
maximum height = (18 m/s)^2 / (2 × 9.8 m/s^2) ≈ 16.53 meters
Therefore, the maximum height reached by the ball is approximately 16.53 meters.
To find the time it takes for a ball dropped from a height of 35 meters to impact the ground, we can use the formula:
time = sqrt(2 × height / acceleration due to gravity)
Given that the height of the building is 35 meters and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the time as:
time = sqrt(2 × 35 m / 9.8 m/s^2) ≈ 2.07 seconds
Therefore, it will take approximately 2.07 seconds for the ball to impact the ground.