ArielT
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Which best describes the speed of a ball as it is thrown straight up into the air and comes back down?
A The ball goes up at a constant speed, stops, then comes down at a constant speed.
B The ball goes up at a constant speed, stops, and increases speed as it comes down.
C The ball goes slower and slower as it goes up, stops, and then goes faster and faster as it comes down.
D The ball goes slower and slower as it goes up, stops, and then comes down at a constant speed.

Sagot :

glendzmorales

Answer:

C. The ball goes slower and slower as it goes up, stops, and then goes faster and faster as it comes down.

Explanation:

Here, we are to describe the speed of a ball thrown straight up and comes back down. According to Newton's law of gravitation, any object thrown upward will fall back because any falling body is under the influence of the earth's gravity and is called free fall.

Therefore, the ball thrown straight upward will eventually slows down because of the pull of gravity until its vertical velocity becomes zero. And when it becomes zero, the only direction to go is to fall down, making an increasingly negative vertical velocity because of the pull of gravity.

Now, let's have some computations for more observations.

1. Let us say, that the ball thrown upward has an initial velocity of 25 m/s. Let us solve for the upward speed with t = 1s, 2s and 3s using the formula

Vf = Vi - gt

where

Vf     is the final velocity at any given time

Vi     is the initial velocity

g      is the acceleration due to gravity, 9.8 m/s²

t       is the given time

Solution:

Vf = Vi - gt

At t = 1s

Vf = 25 - 9.8 (1) = 15.2 m/s    

At t = 2s

Vf = 25 - 9.8 (2) = 5.4 m/s    

At t = 3s

Vf = 25 - 9.8 (3) = - 4.4 m/s    

From this solution, we can say that from the initial speed of the ball thrown straight upward, the speed eventually decreases until it becomes negative which means that the ball goes down. The negative sign means downward direction. In between 2s and 3s is the time where the ball stops in the air. To solve this, we can use:

t = 25/9.8 = 2.551 seconds

2. Now, let's observe the ball going down. The initial velocity is 0 m/s with t= 1s, 2s and 3s using the formula:

Vf = Vi - gt

Solution:

Vf = Vi - gt

At t = 1s

Vf = 0 - 9.8 (1) = - 9.8 m/s    

At t = 2s

Vf = 0 - 9.8 (2) = - 19.6 m/s    

At t = 3s

Vf = 0 - 9.8 (3) = - 29.4 m/s    

Here, we can say that the speed of the ball increases as it goes down.

To learn more, just click the following links:

  • Recommendations about free falling bodies

       https://brainly.ph/question/2162209

  • Additional example with solution and answer

       https://brainly.ph/question/2170448

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