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Answer:
The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second
Explanation:
Horizontal Launch
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
To calculate the vertical distance (y) traveled by the object in terms of the time (t) we use:
[tex]\displaystyle y=\frac{g.t^2}{2}[/tex]
The fastball is thrown horizontally. The distance the ball falls in t=1/4 seconds = 0.25 seconds, is:
[tex]\displaystyle y_1=\frac{9.8*0.25^2}{2}[/tex]
[tex]\displaystyle y_1=0.30625\ m[/tex]
The distance the ball falls in t=1/2 seconds = 0.5 seconds is:
[tex]\displaystyle y_2=\frac{9.8*0.5^2}{2}[/tex]
[tex]\displaystyle y_2=1.225\ m[/tex]
The distance it falls in the second 1/4 second is:
[tex]y_2-y_1=1.225\ m-0.30625\ m[/tex]
[tex]y_2-y_1=0.91875\ m[/tex]
The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second