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

Sagot :

AL2006
-- 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 the
same 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 meters
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.

Answer:

19.6 m

Explanation:

Apex ; good luck to all :)

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