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Need help with projectile motion problem

  1. Apr 27, 2004 #1
    Hi there,
    I'd really, REALLY appreciate a hint for solving the following projectile motion problem.
    "A ball is thrown upward at an angle from a point on a horizontal plane. 10 meters away from that point is a wall. The ball strikes the wall perpendicularly at a height of 4.9 meters. (1) Determine the time from throwing the ball to it hitting the wall. (2) Determine the speed at which it hit the wall. (3) Determine the initial velocity."

    I can't get a handle on this because the initial velocity/angle isn't given. If the ball strikes perpendicularly, that means it's at the top of the trajectory and Vy, or the upward velocity of the projectile, will be 0.0 m/s. Does that sound right? Where do go from here?
     
  2. jcsd
  3. Apr 27, 2004 #2

    Doc Al

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    Staff: Mentor

    what goes up must come down

    If they don't give it, that generally means you don't need it. (Or that you can figure it out.)
    Right! Now think about the motion in the y-direction only. If you toss a ball up, and it reaches a maximum height, how long does it take to do it? If that puzzles you, what about the reverse problem: If you drop a ball from a certain height, how long does it take to fall? Get the connection?
     
  4. Apr 27, 2004 #3
    Oh! I kind of know this kind of problem! That 4.9 m business, that's how far something falls in one second with the gravitational constant of 9.8m/s/s. What goes up, must come down, so was it going maybe 10 m/s or something? It went 10 meters in one second?
     
  5. Apr 27, 2004 #4

    Doc Al

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    Staff: Mentor

    Go, Holly! Absolutely correct.
    Right again. V = gt = 9.8 m/s. That's the ball's initial speed in the y-direction.
    Oops. It went 4.9 meters in one second. Remember? :wink:

    The 9.8 m/s is the ball's instantaneous speed (in the y-direction) at the moment it is thrown. The average speed (in the y-direction) as it goes up to its maximum height is 4.9 m/s. Got it?
     
  6. Apr 27, 2004 #5
    Got it! (slaps his forehead). I kept thinkingi that i couldn't use "y = v(initial y) x t + 0.5 x g x t(squared)" because i didn't know v(initial y) and t. Now i see that the vertical height of 4.9 m is not just any height, but a very special one in this case, in the sense that only one v(initial y) could achieve it. Thanx
     
  7. Apr 27, 2004 #6
    Projectile motion

    This problem is, unfortunately, contrived because of the special values used for the numerical properties. Here is how you solve ANY simple projectile motion problem.

    While the value of the initial speed and angle are not given, we know that they must have definite values. So call them [tex]v_0[/tex] and [tex]\theta[\tex] and pretend you know them.

    Now, picture a fluorescent light fixture that hangs over the entire system, shining down. The light will cast a shadow of the ball on the ground, right? What is the motion of this shadow?

    Well, the shadow will move across the ground at constant speed for 10 meters. How fast does this shadow travel? That's easy, [tex]v_x = v_0\cos\theta[/tex]. Since the shadow moves across at constant speed, its distance/velocity/time relationship is simply

    [tex] 10 = (v_0\cos\theta)t[/tex]

    Now, pretend you can cast a vertical fluorescent light on the ball, printing an image of its shadow on a wall. This shadow is like an object in free fall. It travels initially upwards at a speed [tex] v_{oy} = v_0 \sin\theta[/tex], but slows at a rate of [tex]g = 9.8[/tex] (in SI units). Its equation of motion must therefore be

    [tex] 4.9 = (v_0\cos\theta)t + \frac{1}{2}(-9.8) t^2[/tex]

    Now we know one more bit of information about the system -- its final velocity in the vertical direction is 0. Using the second equation of motion we get

    [tex] 0 = 4.9 + (-9.8)t[/tex]

    Three equations, three unknowns. The time from the last equation must be [tex]t = 0.5[/tex] seconds. Substitute this value into the first two equations and solve the resulting two equations, two unknowns.

    The beauty is this procedure will solve any simple projectile motion problem a science prof can throw at you.
     
  8. Apr 27, 2004 #7
    Projectile motion

    This problem is, unfortunately, contrived because of the special values used for the numerical properties. Here is how you solve ANY simple projectile motion problem.

    While the value of the initial speed and angle are not given, we know that they must have definite values. So call them [tex]v_0[/tex] and [tex]\theta[/tex] and pretend you know them.

    Now, picture a fluorescent light fixture that hangs over the entire system, shining down. The light will cast a shadow of the ball on the ground, right? What is the motion of this shadow?

    Well, the shadow will move across the ground at constant speed for 10 meters. How fast does this shadow travel? That's easy, [tex]v_x = v_0\cos\theta[/tex]. Since the shadow moves across at constant speed, its distance/velocity/time relationship is simply

    [tex] 10 = (v_0\cos\theta)t[/tex]

    Now, pretend you can cast a vertical fluorescent light on the ball, printing an image of its shadow on a wall. This shadow is like an object in free fall. It travels initially upwards at a speed [tex] v_{oy} = v_0 \sin\theta[/tex], but slows at a rate of [tex]g = 9.8[/tex] (in SI units). Its equation of motion for displacement, [tex]\vec{d} = \vec{v}_0t + \frac{1}{2}\vec{a}t^2[/tex], must therefore be

    [tex] 4.9 = (v_0\cos\theta)t + \frac{1}{2}(-9.8) t^2[/tex]

    Now we know one more bit of information about the system -- its final velocity in the vertical direction is 0. Using the equation of motion for velocity [tex]\vec{v} = \vec{v}_0 + \vec{a}t[/tex] we get

    [tex] 0 = 4.9 + (-9.8)t[/tex]

    Three equations, three unknowns. The time from the last equation must be [tex]t = 0.5[/tex] seconds. Substitute this value into the first two equations and solve the resulting two equations, two unknowns.

    The beauty is this procedure will solve any simple projectile motion problem a science prof can throw at you.
     
    Last edited: Apr 27, 2004
  9. Apr 27, 2004 #8
    Question:
    Regarding [tex] 10 = (v_0\cos\theta)t[/tex]
    Although we know that the shadow has moved 10 meters, how do we know from the start that it is moving across the ground at 10 m/sec? Did you skip a step?
     
  10. Apr 28, 2004 #9

    Doc Al

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    Staff: Mentor

    The only thing contrived about this problem is that the ball hits the wall perpendicularly. (Lot's of luck getting that to happen. :smile: ) But the height is arbitrary--- if it was 8.3m instead of 4.9m, you'd solve it the same way. Of course you wouldn't get a neat answer of 1 second. (So I guess that's contrived. But it doesn't change how you solve it.)
     
  11. May 4, 2004 #10
    The shadow does not move across the ground at 10 m/s, but at [tex]v_o\cos\theta[/tex] m/s.

    The original speed of the object is denoted [tex]v_0[/tex]. The speed of the shadow on the ground is the horizontal component of the initial velocity vector, which is [tex]v_0\cos\theta[/tex].

    That is why [tex]10=v_0\cos\theta[/tex], because the equation of motion for distance is [tex] d = vt[/tex] when the velocity is constant.
     
  12. May 4, 2004 #11
    RE: "Well, the shadow will move across the ground at constant speed for 10 meters"

    I think you misread my statement. I did not say that the speed was 10 meters/second.
     
    Last edited: May 4, 2004
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