PHYSICS QUESTION, can you please help me; its in the details.
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PHYSICS QUESTION, can you please help me; its in the details.

[From: ] [author: ] [Date: 11-10-24] [Hit: ]
0 m high and 130.0m away and the ball is hit when it is 1.00 m above the ground, do the player hit a homerun.-Ill reserve the 1m. for now.......
a baseball hits a ball with an initial velocity of 35.3 m/s at an angle of 35.0 degrees. If the outfield wall is 21.0 m high and 130.0m away and the ball is hit when it is 1.00 m above the ground, do the player hit a homerun.

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I'll reserve the 1m. for now.
Vertical velocity component of hit = (sin 35) x 35.3, = 20.25m/sec.
Horizontal component = (cos 35) x 35.3, = 28.9m/sec.
Time for ball to reach wall = (130/28.9), = 4.498 secs.

Time for ball to rise to max. height = (v/g), = (20.25/9.8), = 2.066 secs.
Maximum height attained = (v^2/2g), = (20.25^2/19.6), = 20.92 metres.
Add the 1m. now = 21.92m.
Distance to max. height = (2.066 x 28.9) = 59.7 metres horizontally.
Distance left to wall = (130 - 59.7) = 70.3 metres.
Time to drop in 70.3m. = (70.3/28.9) = 2.43 secs. That checks with (4.498 - 2.066) = 2.432 secs.
Distance ball drops in 2.43 secs = 1/2 (t^2 x g), = 28.81 metres.

The ball is going to reach ground before it reaches the wall, so no home run.
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