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anonymous

  • 4 years ago

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 18.0m/s at an angle of 37.5 above the horizontal. A) What is the horizontal component of the ball's velocity just before it is caught? B) How long is the ball in the air? Can someone please explain to me how this works not just the answer I'm somewhat confused.

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  1. TuringTest
    • 4 years ago
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    Break the motion vector into components. Here are the accelerations:\[a_x=0\]\[a_y=-g\]Horizontal velocity is constant because no force acts in that direction. The vertical component has the force of gravity acting on it. Integrating our acceleration components we get the velocity.\[V_x=|V|\cos\theta\]\[V_y=|V|\sin\theta-gt\]Where |V| is the magnitude of the initial velocity (given in the problem) and theta is the angle it was thrown at. Let us call the time when the ball is at the highest point of its trajectory t'. We know that at that point the vertical velocity component is zero, so we can use that to find t':\[0=|V|\sin\theta-gt'\to t'=\frac{|V|}g\sin\theta\]It will take the same amount of time to fall back to its original height, so the total time of flight will be 2t'. Note: the total time of flight will be 2t' ONLY if the object returns to its original height.

  2. anonymous
    • 4 years ago
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    I thought so! thank you!

  3. anonymous
    • 4 years ago
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    I got 1.12 which is half of the answer so just needed to double it! thank you!

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