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During a baseball game, a batter hits a high .
pop-up. If the ball remains in the air for 6.0 s,
how high does it rise? Hint: Calculate the
height using the second half of the trajectory.

Respuesta :

Answer: Due that we don't know the initial speed after hitting the ball, we are going to accept that the ball goes up for half of the time and then falls during other half part, that is 3.0 seconds each. Then we know that ball's movement is ruled by the acceleration of gravity formula, as follows: H = Vi * T + 1/2 * g * T^2 V = Vi + g * T where: H is height, Vi initial speed, g gravity acceleration and T time When we only consider the second half of the trajectory, we have that initial speed at the top of that movement is zero, because ball goes up till top, where stops and starts to go down, so : H = 0 * 3 + 1/2 * 32 * 3^2 = 144 ft. So the height of the pop-up is 144 feet.

kapoorprachi783

The ball rise 144 feet high.

How to find the height of the ball?

Due to that, we don't know the initial speed after hitting the ball, we are going to accept that the ball goes up for half of the time and then falls during another half part, which is 3.0 seconds each.

Then we know that the ball's movement is ruled by the acceleration of gravity formula, as follows: H = Vi * T + 1/2 * g * T^2 V = Vi + g * T

Where: H is height, V is initial speed, g is gravity acceleration, and T is time.

When we only consider the second half of the trajectory, we have that initial speed at the top of that movement is zero, because the ball goes up to the top, where stops and starts to go down, so:

H = 0 * 3 + 1/2 * 32 * 3^2 = 144 ft.

So the height of the pop-up is 144 feet.

Height is a measure of vertical distance, either vertical extent or vertical position

Learn more about height here: brainly.com/question/6613758

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