Physics  check
posted by Anonymous on .
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

It looks OK but I don't think you meant to say
"Note: maximum height will be 0, right?"
The vertical VELOCITY component at maximum height is zero.
With all those(4)significant figures, why are you using 10 for g instaed of 9.80 m/s^2 ? Unless they tell you do do that, I'd use a more accurate value. With four figures, it is about 9.804, but varies from place to place. 
wooops I meant to say ask if the vertical velocity at maximum height is zero. And you answered that. Thanks! Yes, we were told to use 10 instead of 9.8.

I'm not sure if I did a right
but V2 x t
= 30.41 x 2
= 60.82m 
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
Wait...it says location. How do I calculate the location? I only calculated it's vertical velocity. 
The horizontal location at time t is
X = V1parallel * t
Th vertical location is V1perpendicular* t  (g/2)t^2 
thanks!
d) is wrong, too. Can you help me with that?