A home run is hit in such a way that the baseball just clears a wall 20.0 m high, located 114 m from home plate. The ball is hit at an angle of 33.0° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) (a) Find the initial speed of the ball. Incorrect: Your answer is incorrect. Write the expressions for the horizontal and vertical positions as functions of time. Then use the equations to find the initial speed and the time when the baseball reaches the height of interest. m/s

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aristocles aristocles · Answer 1 · 2019-10-02T06:23:55+02:00

Answer:

[tex]v_o = 40.56 m/s[/tex]

Explanation:

Let the initial speed of the ball is given as

[tex]v_i = v_o[/tex]

then angle of inclination is given as 33 degree

now its two components of velocity is given as

[tex]v_x = v_o cos33[/tex]

[tex]v_y = v_o sin33[/tex]

now the two positions with time given as

[tex]y = y_o + v_y t + \frac{1}{2}at^2[/tex]

[tex]y = 1 + v_osin33 t - 4.9 t^2[/tex]

[tex]x = v_o cos33 t[/tex]

now when it hit the wall so we will have

[tex]x = 114 = v_o cos33 t[/tex]

[tex]y = 20 = 1 + v_o sin33 t - 4.9 t^2[/tex]

now from above two equations we will have

[tex]19 = \frac{114}{v_o cos33}(v_o sin33) - 4.9 t^2[/tex]

[tex]19 = 74 - 4.9 t^2[/tex]

[tex]t = 3.35 s[/tex]

now from above equations again

[tex]114 = v_o cos33 (3.35)[/tex]

[tex]v_o = 40.56 m/s[/tex]