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Projectile thrown from car

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A car is driving 40 m/s north on a flat surface when the driver throws a ball out the window. The ball leaves his hand moving horizontally and perpendicular to the car with a speed of 20 m/s with respect to the car.

    At the moment the ball leaves the driver's hand, it is 1m above ground. Find the total distance traveled by the ball between its release and the instant it hits the ground. Neglect air resistance or any other forces other than the weight of the ball.

    2. Relevant equations
    Kinematics:

    y = Vot + (1/2)at2
    Vf = Vo + at
    Vf2 = Vo2 + 2ay
    y = t * [ (Vo + Vf) / 2 ]

    Gravity is 9.82 m/s^2

    3. The attempt at a solution

    Vector algebra says the initial velocity of the ball is sqrt(40^2 + 20^2) = 44.7 m/s

    The final velocity will be 0 m/s I believe.

    I've tried a bunch of weird things with Kinematics but I keep getting results that are way off from what is reality. Projectile motion is usually not that difficult for me but that's because I always have the time. In this case I don't so I have to solve for time before doing much else.

    y = Vot + (1/2)at2

    y = 44.7t - 4.91 m/s2 * t2

    Without knowing distance I can't solve for the time here.

    Using Vf = Vo + at gave me a weird answer that I knew was wrong. It won't take over 4 seconds to hit the ground I'd think.

    Not sure where to go from here...
     
    Last edited: Jan 29, 2014
  2. jcsd
  3. Jan 29, 2014 #2

    collinsmark

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    That's right! :approve: Keep that value handy because you'll use it later. :smile: That said, you might want to hold on to some more significant figures for now. You can always reduce the number of significant figures in your final answer. In other words, I wouldn't round to 44.7 m/s so soon. Keep more significant figures for now (like 44.721 m/s, for example).

    Ummm, no. The ball/rock will hit the ground with a non-zero velocity. It definitely won't be zero. If it helps, think of it as the ball/rock's velocity at the instant before it hits the ground.

    It doesn't really matter for this problem though. You can find the final answer without ever calculating the ball/rock's final velocity.

    That's a handy equation and you'll use it for this problem. As a matter of fact, you'll use it for your next step.

    [Edit: The use of the equation isn't right. You're mixing up your components. You're mixing up the velocity of one direction with the acceleration in another.]

    That particular kinematics equation is not so useful for this particular problem.

    Go back to your
    y = v0t +(0.5)at2
    equation.

    But before using it, remember to break up the motion of the ball/rock into its components. You've already combined the two horizontal components of velocity, which is fine, and we'll come back to that. But does acceleration due to gravity act on the ball/rock in any of these horizontal directions?

    What about the vertical component (up-down component) of the ball's motion. What is the initial up-down velocity of the ball/rock?

    What about acceleration due to gravity? Does the gravitational acceleration act on the ball/rock in the up/down direction?
     
    Last edited: Jan 29, 2014
  4. Jan 29, 2014 #3
    Ah, because we know that the distance from top to bottom is 1m and its acceleration is gravity, time can be found from that.

    1m = 4.91 t^2
    .451s = t

    And that time is universal for all components. So I can solve for the distance of the x-component (assuming x-axis is east to west). Acceleration for the x component is zero.

    dx = 20m/s * .451s = 9.02m

    If the Z-axis is up and down, Y-axis South to North, X-axis East to West, what would the acceleration of the Y component be? The car is traveling north and throws to the west, is this when I use the 44.721 m/s?
     
  5. Jan 29, 2014 #4

    collinsmark

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    That looks right to me! :approve: Very nice.

    Again though, I'd recommend keeping some more signification figures for these intermediate steps.

    That's not quite what I had in mind calculating the x- and y- distances individually (since you already combined their velocities), but that's okay. Calculating the x- and y- distances individually and then combining them later will also work.

    So you've found the horizontal distance the ball/rock traveled that is perpendicular to the car's motion.

    What's the horizontal distance of the ball/rock that's parallel to the car's motion?

    Gravitational acceleration only matters in the up-down direction, and you've already taken care of that (in this case, the z-direction).

    There isn't any gravitational acceleration acting in the north-south direction, nor is there in the east-west direction. That's because these are both horizontal directions. Projectile motion components are at constant velocities in these directions.

    You could, yes!

    There's more than one way to do this problem, by the way. One way is to combine the constant, horizontal velocities, and then solve for the final distance. The other way is to calculate the individual, horizontal component distances, and then combine the horizontal, distance components as the last step.
     
  6. Jan 30, 2014 #5
    Alright I think I know what you mean. 44.721 m/s is the initial velocity for the horizon in general right? So it already combined the x and y axis.

    So
    distance of horizon = 44.721m/s * .451s = 20.169m

    So it travels 20169m in the NW direction and 1m down. This would be sqrt(20.169^2 + 1^2) = 20.194m is the distance traveled.

    Does this sound correct to you?
     
  7. Jan 30, 2014 #6

    collinsmark

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    Yes, that's exactly the right idea! :approve:

    I almost would have accidentally forgotten to even include the vertical distance, but the problem statement did say "total distance," so good job! :smile:

    I got a slightly different answer (very small difference) due to rounding differences, by the way. But it's very close to your answer.
     
  8. Jan 30, 2014 #7
    Awesome thank you so much! I never dealt with projectile problems that involved a third axis/vector. But now I can. Makes me smile :)
     
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