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

Physics
Tutor: None Selected Time limit: 1 Day

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

Thank you for the opportunity to help you with your question!

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

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Oct 4th, 2015
...
Oct 4th, 2015
Dec 6th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer