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

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?
Respond to this Question
Similar Questions

physics
You throw a ball toward a wall with a speed of 25m/s and at an angleof 40 degree above the horizontal. The wall is 22m from the realease point of the ball. a) How far above the release point does the ball hit the wall? 
physics
A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string. 1)Determine the Mass … 
Physics  check
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. … 
Physics
An object is projected at an angle of 53.1degrees above the horizontal at a speed of 50m/s. Sometime later it enters a narrow tube positioned at an angle of 45degrees to the vertical. Determine: a) The initial horizontal and vertical … 
Physics
An object is projected at an angle of 53.1degrees above the horizontal at a speed of 50m/s. Sometime later it enters a narrow tube positioned at an angle of 45degrees to the vertical. Determine: a) The initial horizontal and vertical … 
Physics  Inelastic Collision ( check + help)
A billiard ball of mass 0.55kg moves with a velocity of 12.5m/s towards a stationary billiard ball of the identical mass and strikes it with a glancing blow. The first billiard ball moves of at an angle of 29.7deg clockwise from its … 
Physics
A ball rolls on a circular track of radius 0.65 m with a constant angular speed of 1.2 rad/s in the counterclockwise direction. Part A If the angular position of the ball at t= 0 is theta= 0, find the x component of the ball's position … 
Physics
Hypothetically, you are standing on the balcony of an apartment on the 10th level of a building and you throw a ball at some angle,è , above the horizontal away from the building. the ball lands on the ground 10m from the building, … 
Physics
HELP PLEASE. Hypothetically, you are standing on the balcony of an apartment on the 10th level of a building and you throw a ball at some angle, 0 , above the horizontal away from the building. the ball lands on the ground 10m from … 
AP PHYSICS
A golf ball is hit with an initial velocity of 50m/s at an angle of 25° above the horizontal. A) How far does the ball travel?