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A second baseman tosses the ball to a first baseman, who catches it a the same level from which it was thrown. The throw is made with an initial speed o 19.0 m/s at an angle of 37 degrees above the horizontal. Neglect air resistance. How long is the ball in the air?

Sagot :

AL2006
-- The toss was directed 37° above horizontal, at 19 m/s.

The horizontal component is 19 cos(37°) = 15.174 m/s .

The vertical component is 19 sin(37°) = 11.434 m/s .

The ball will continue to climb until its vertical speed is zero,
then will begin falling.  It'll fall to the same level from which
it was thrown in the same time it spent rising.

Time spent rising = (11.434 m/s) / (acceleration of gravity) = 11.434/9.8 = 1.166 second.

Return to the same level from which it was thrown in (2 x 1.166) = 2.332 seconds.

The ball is in the air for 2.332 seconds.

In that amount of time, it travels horizontally (2.332) x (15.174) = 35.39 meters.
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