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Easy Physics problem need help

  1. Nov 1, 2003 #1
    Ok, this is a two part problem.

    Part I: We have a ball at the top of a roof that has a 30 degree angle at the horizontal. We let the ball roll and it takes 2.55 seconds to reach the bottom of the 6.34 meter roof. The ball weighs .35 kg. **DISREGARD ANY FRICTION** I.E friction of the ramp, wind resistance, any type of friction. The main part of this is to find the Final Velocity at the end of the 6.34 meter roof. So you have Vi(initial velocity) = 0, Vf(Final Velocity) = ????, G(gravity) = 9.8m/s^2. The ramp has a 30 degree angle at the horizontal, and it takes 2.55 seconds to get to the bottom.

    Part II: After we find the Final Velocity, we have to figure out how far the ball will travel in the air. After it reaches the final velocity, it falls off the roof 2.51 meters to the ground. The *main part* of the entire problem is to figure out how far it will go off the roof. The Y axis is 2.51 meters, and we need to find out what the X axis will be. We need to find out Dx(distance of x), Vx(Velocity of x), and T(time it takes to reach the ground).

    My friends and I have been having trouble with this problem all day. We keep coming up with wrong, impossible answers. Any help is appreciated. Thanks.
     
  2. jcsd
  3. Nov 1, 2003 #2
    First of all a ball cannot weigh .35kg, it has a mass of .35kg. Thats like saying the distance from A to B is 57Volts.

    The first step to solving ur problem is finding the resultant force acting on the ball. This can be done by drawing one of those cross-hair things. I will do my best to demonstrate how, using only characters on the ASCII pallette.

    *
    *
    o*
    ...
    | ...
    | ...
    | ...

    ... = roof
    o = ball
    | = Force of weight =mass x accel due to grav= 0.35 x 9.8 = 3.43
    * = The normal reaction force of the roof (acting perpendicular to the roof)

    If you place a crosshair on the system so the the middle is on the body and one line is parallel to the normal force of the roof and the other is offset 30 degrees clockwise to the force of weight.

    You can say that the forces acting perpendicular to the roof are ballanced since there is no change in velocity in a perpendicular direction with respect to the roof. Taking the perpendicular componant of the Weight using trig would be 3.43Cos30 and therefore the normal force would be 3.43Cos30. This means that the resulatant force ends up going down the roof 3.43Sin30 = 1.175N

    Put that into F=ma (resultant force = mass x accel) to get

    1.715 = 0.35a therefore a = 0.6ms^-2 down the roof

    substitute this value into v=u+at to get v=0.6 x2.55 = 1.53

    DO THE REST UR SELF. Maybe i have helped u enuf to do so

    it has been an honour to answer ur question. If u need more help chat 2 me on msn on mceddy2001@hotmail.com. Soz about the blabber but i cant draw the diagrams on here so i need 2 explain more. If u want me 2 draw a better diagram in paint i will be happy providing i have the time
     
  4. Nov 1, 2003 #3
    sorry my diagram which i spend ages on got ruined by the system. Many apologies it seems to have taken out all of the necessary spaces.
     
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