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.

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.

The maximum height reached by the baseball is 44.1 m

The given parameter:

time for the entire motion of the baseball = time spent in air, T = 6.0 s

time to reach the maximum height,

To find:

the maximum height traveled by the baseball,

The maximum height reached by the baseball using second half of the trajectory is calculated as;

where;

u is the initial velocity = 0

Thus, the maximum height reached by the baseball is 44.1 m

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Iwas going to ask that quesytion  sorry dude i dont  know : (    : /