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A ball is thrown horizontally from the top of a 24.7 m building with a speed of 13.2m/s. Assuming level ground, the ball lands ___ m from the base of the building.

Sagot :

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Answer:

The horizontal distance traveled by the ball is 29.7 m.

Explanation:

Given;

height of the building, h = 24.7 m

initial horizontal velocity, Vₓ = 13.2 m/s

The time taken for the ball to fall from the vertical height is calculated as;

[tex]h = V_yt + \frac{1}{2} gt^2[/tex]

where;

[tex]V_y[/tex] is initial vertical velocity = 0

[tex]h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 24.7}{9.8} } \\\\t = 2.25 \ s[/tex]

The horizontal distance traveled by the ball is calculated as;

[tex]X = V_x t\\\\X = 13.2 \times2.25\\\\X = 29.7 m[/tex]

Therefore, the horizontal distance traveled by the ball is 29.7 m.

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