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

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

I'm not sure if I did a right

but V2 x t
= 30.41 x 2
= 60.82m

Answer 5

The horizontal location at time t is
X = V1parallel * t

Th vertical location is V1perpendicular* t - (g/2)t^2

Answer 6

thanks!

d) is wrong, too. Can you help me with that?