You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

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Accepted Answer

-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s-- The acceleration of gravity is 9.8 m/s².-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for thesame amount of time before it hits the ground.-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds==================================-- The horizontal component of the ball's velocity is  14 cos(51°) = 8.81 m/s-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = 19.56 metersbefore it hits the ground.As usual when we're discussing this stuff, we completely ignore air resistance.

Suggested Answer

Answer:19.6 mExplanation:Apex ; good luck to all :)

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