Physics, 14.09.2019 08:30 kaywendel2008

Abasketball center holds a basketball straight out, 2.0 m above the floor, and releases it. it bounces off the floor and rises to a height of 1.5 m. a) what is the ball's velocity just before it hits the floor? b) what is the ball's velocity just after it leaves the floor? c) if the ball is in contact with the floor for 0.02 seconds, what are the magnitude and direction of the ball's average acceleration while in contact with the floor?

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Answer from: mima851

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

Answer from: Quest

Neptune is the eighth planet from the sun. it was the first planet to get its existence predicted by mathematical calculations before it was actually seen through a telescope on sept. 23, 1846. irregularities in the orbit of uranus led french astronomer alexis bouvard to suggest that the gravitational pull from another celestial body might be responsible. german astronomer johann galle then relied on subsequent calculations to spot neptune via telescope. previously, astronomer galileo galilei sketched the planet, but he mistook it for a star due to its slow motion. in accordance with all the other planets seen in the sky, this new world was given a name from greek and roman mythology — neptune, the roman god of the sea.

Answer from: Quest

The answer would be a) -80of

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