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A football player kicks a ball with an initial velocity of 30 meters per second at an angle of 53° above the horizontal. how long will the ball be in the air for

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At time t, the ball attains a height y of

y = (30 m/s) sin(53º) t - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 0:

(30 m/s) sin(53º) t - 1/2 g t² = 0

t ((30 m/s) sin(53º) - (4.90 m/s²) t ) = 0

t = 0 or (30 m/s) sin(53º) - (4.90 m/s²) t = 0

We ignore the first solution, since that refers to the moment the ball is kicked.

(30 m/s) sin(53º) - (4.90 m/s²) t = 0

(30 m/s) sin(53º) = (4.90 m/s²) t

t = (30 m/s) sin(53º) / (4.90 m/s²)

t ≈ 4.9 s

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