The distance traveled by the ball before hitting the ground is 7.2 m.
The given parameters:
The time of motion of the ball is calculated as follows;
[tex]h= v_y_i t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2 \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 1}{9.8} }\\\\t =0.45 \ s[/tex]
The horizontal distance traveled by the ball before hitting the ground is calculated as follows;
[tex]X = v_x \times t\\\\X = vcos\theta \times t\\\\X = 20 \times cos(37) \times 0.45\\\\X = 7.2 \ m[/tex]
Thus, the distance traveled by the ball before hitting the ground is 7.2 m.
Learn more about horizontal distance here: https://brainly.com/question/24784992
