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

-- 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.

Answered by AL2006 | 2024-06-10

19.6 m ;

Answered by dsiwjfie | 2024-06-13

The ball lands approximately 19.56 meters away from its initial position when kicked with a speed of 14 m/s at an angle of 51°. This result is derived by analyzing the vertical and horizontal components of the motion. The calculations assume no air resistance.
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Answered by AL2006 | 2024-10-11