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Homework Help: Projectile Motion at an Angle

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data
    You throw a ball. The ball leaves your hand with a velocity of 15 m/s at an angle 60 degrees above the horizontal. Acceleration due to gravity is 9.8 m/s/s.
    a) Determine the velocity (magnitude and direction) of the ball at the maximum height.
    b) How much time does it take for the ball to reach maximum height?
    c) Determine the maximum height.

    2. Relevant equations
    (I know how to draw the diagram, so that part is good)
    I think the components of the velocity at maximum height can be found using these equations:
    a) VoY= Vosin60 = 15m/s(sin60) = 12.99 m/s velocity of ball in vertical direction (initial)
    VoX= Vocos60 = 15 m/s(sin60) = 7.5 m/s velocity of ball in horizontal direction (initial)
    Since there is no vertical velocity at max height, then the velocity at max height is just the magnitude of the horizontal component: 7.5 m/s horizontal direction??????

    b) Time = Vy= VoY + ayt , set Vy = 0 at Y max.
    0 = 12 m/s + (-9.8 m/s/s)t
    t = (Vfy - Voy)/ -9.8m/s/s = (0 - 12.99)/-9.8 m/s/s = 1.32 seconds

    c) Ymax. = Voyt + 1/2at^2 = (12.99 m/s)(1.32 sec.) + 1/2(-9.8m/s/s)(1.32 sec.)^2

    = 8.60 meters max height

    3. The attempt at a solution
    see above. im not sure if i did this problem correctly, or if i understand it quite. there are some things about it that i understand, others not so much.
    I think the velocity at max height would just be the component 7.5 m/s in the horizontal direction, since suposedly, there is no vertical velocity at max height. Is this assumption correct? Have i used the correct equations and in the right context?? If not, please help me to understand this type of problem better.
    I understand that at max height, there is only the force of gravity 9.8 m/s/s downward, the horizontal velocity for a projectile is constant, and there is no horizontal acceleration. I assume there is no horizontal acceleration, because the horizontal veloc. remains constant?
    However, if there is no vertical velocity at max height then why is the object still accelerating at 9.8 m/s/s downward due to gravity??? It seems like a contradiction, something cant be accelerating due to some force and not also have a velocity. Increasing velocity is the definition of acceleration, so there would be vertical velocity at max height. There is always vertical velocity due to the force of gravity accelerating the object downward.
  2. jcsd
  3. Jan 5, 2010 #2
    Correct. The magnitude of the the two velocity vectors are as follows

    sqr[ (7.5)^2 + (0)^2 ] = 7.5

    Everything in your solution looks fine to me.

    Because firstly the object the object is moving in the positive direction but accelerates towards the negative direction. Eventually its going to have to cross the point of 0m/s in order to start moving in the negative direction instead of the positive direction.

    Think of a thermometer, If its 20 degrees outside and it drops to -10 degrees it eventually crossed 0 degrees right?
    Last edited: Jan 5, 2010
  4. Jan 5, 2010 #3
    i understand. thanks!
  5. Jan 5, 2010 #4
    quick question, just to make sure i get this: the velocity for the ball at max height is the initial velocity component equation because the velocity for horizontal motion remains constant, am i right?? And the initial vertical velocity component does not apply to the first question (only with regard to finding time at max height using this equation Vy = Voy + ayt)
    When i had first done this problem, i had confused the two initial component equations, not realizing this. i had tried to incorporate the vertical veloc. component equation with the first question even though i know that there is no vertical velocity. anyhow, now i know that i should have written the vertical component equation down near the second question(in my homework paper) to not confuse myself with the horizontal component equation. Both of them are found using a very similar equation. Do you see what i mean how I would confuse the two??
  6. Jan 5, 2010 #5
    I think i had confused the two equations because i didn't understand why the time equation
    t = Vfy - Voy / -9.8 m/s/s ) uses the vertical component instead of the horizontal component. is there a reason for this??
  7. Jan 5, 2010 #6
    In other words, if there is no vertical velocity, why must i use a vertical veloc. component to find the time at max height. If vertical velocity is non-existent, why not use the horizontal veloc. component to find the time instead??
  8. Jan 5, 2010 #7
    You're thinking that there is NO vertical component of velocity and therefore its impossible to solve for time when there is NOTHING to work with right?

    You need to understand when we say there is no vertical component of velocity we mean Vy = 0.

    And with that fact we are able to solve for time.
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