a = ( V2 - V1)/( t2 - t1)
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
-- The vertical component of the ball's velocity is 14 sin(<span>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
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-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
B would be the correct answer
