Physics, maximum height, distance, and horizontal/vertical components
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Physics, maximum height, distance, and horizontal/vertical components

[From: ] [author: ] [Date: 11-09-13] [Hit: ]
height attained = (v^2/2g), = (4^2/19.6), = 0.82m.Time to reach max.......
A man kicks a ball from ground level with an initial velocity of 8.0 m/s at an angle of 30 above the horizontal.

What is the maximum height of the ball above the ground?

How far from the man will the ball land?

What are the horizontal and vertical components of the ball's velocity half a second into the flight?

-I worked the first problem and somehow end up with .82 meters, which I don't think is correct.

-
Vertical component of kick velocity = (sin 30) x 8, = 4m/sec.
Horizontal component = (cos 30) x 8, = 6.928m/sec.
a) Max. height attained = (v^2/2g), = (4^2/19.6), = 0.82m.
Time to reach max. height = sqrt. (2h/g), = 0.4091 secs. Double it, = 0.8182 secs.
b) Distance = (0.8182 x 6.928), = 5.67 metres.

c) Horizontal component = still 6.928m/sec. Vertical component = 4 - (9.8 x 0.5) = 0.9m/sec., headed down.
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