Physics  check
posted by Anonymous .
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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