During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 5.1 s, how h

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 5.1 s, how high does it rise? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

Oct 4th, 2015

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The height of the ball at time is

h(t) = h(0) + v(0)t - (1/2)gt²

Since the initial height of the ball, h(0), is not given, we have to assume that the height of the ball at t = 5.38 seconds is equal to its initial height.

Then we can solve for v(0):

h(t) = h(0) + v(0)*t - (1/2)(9.8)t²

h(t) - h(0) = 0 = v(0)*(5.1) - (1/2)(9.8)(5.1)²

v(0)*(5.1) = (1/2)(9.8)(5.1)²

v(0) = (1/2)(9.8)(5.1)

v(0) = 24.362 m/s


The equation for the height of the ball is then:

h(t) = h(0) + 24.362t - (1/2)(9.8)t²


The maximum height of the ball then occurs at:

t = -(24.362) / (2*(-1/2)(9.8) = 2.32


The height that the ball has risen is then:

Height risen = h(t) - h(0) = v(0)t - (1/2)(9.8)t²

Height risen = (26.362)(2.32) - (1/2)(9.8)(2.32)²

Height risen approx = 33.46 m



Oct 4th, 2015

Modify height it is 5.1 m 

Oct 4th, 2015

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