(a) The time taken for the ball to reach the height of the fence is 4.1 s.
(b) The position of the ball from the batter at the time is 111.33 m.
(c) Henry will have a home run.
The time taken for the ball to reach the height of the fence is calculated as follows;
[tex]h = v_0_y t - \frac{1}{2} gt^2\\\\(3 - 1.2) = (34 \times sin37)t - \frac{1}{2} (9.8)t^2\\\\1.8 = 20.46t- 4.9t^2\\\\4.9t^2 - 20.46t+ 1.8 = 0\\\\a = 4.9, \ b = -20.46, \ c = 1.8\\\\t = \frac{-b\pm \sqrt{b^2 - 4ac} }{2a} \\\\t = 4.1 \ s[/tex]
The position of the ball at the time is calculated as follows;
[tex]X = v_x t\\\\X = (34 \times cos37) \times 4.1\\\\X = 111.33 \ m[/tex]
Since, 111.33 m is greater than 106 m, Henry will have a home run.
Learn more about about time of motion here: https://brainly.com/question/2364404
