Posted by **Anonymous** on Thursday, March 13, 2008 at 11:03am.

A ball is thrown at a speed of 39.05m/s at an angle of 39.8degrees above the horizontal. Determine:

V1perpendicular component:

V1Sin(theta)

= 25m/s

V1paralell component:

V1Cos(theta)

= 30m/s

a) the ball's location 2s after being thrown.

V2paralell = V1paralle = 30m/s

V2perpendicular = V1+at

= 25+(-10)(2)

= 5m/s

V2 = sqrt of 30^2 + 5^2

= 30.41m/s

b) the time it takes to reach maximum height.

t = (V2perp - V1perp)/a

= -25/-10

= 2.5s

Note: maximum height will be 0, right? It stops momentarily before it goes down.

c) the maximum height above ground.

V2^2perp = V1^2perp + 2ad

= 25^2 + 2(-10)d

-625 = -20d

-625/-20 = d

d = 31.25m

d) the position and velocity 3s after being thrown.

V2perp = V1perp + at

= 25 + (-10)(3)

= -5m/s

d = v1perp(t) + 1/2at

= (25x3) + .5(-10)(3)

= 60m

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