Projectile Questions: Solving Max Height & Xmax

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The discussion revolves around calculating the maximum height and horizontal distance (xmax) for two tennis ball toss experiments. For the far toss, the initial height is 1.3m, with a flight time of 1.92 seconds and a total distance of 23.4m. The formula delta y = VyoT - 0.5gt^2 is used to find maximum height, where Vyo is the initial vertical velocity, which can be derived from the known parameters. The horizontal motion is constant, allowing for the calculation of xmax by relating the time to reach maximum height with the total flight time. The conversation emphasizes understanding vertical and horizontal motion to solve for the required values accurately.
  • #31
Yes they do. And because the motion is not accelerated, how does ## \frac {s_1} {t_1} ## compare with ## \frac {s_2} {t_2} ##?
 
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  • #32
They are both constant?
 
  • #33
Of course they are constant, because they are ratios of constants. But that does not answer the question at all.
 
  • #34
Oh sorry I meant that the velocities are constant, but other than that I'm not sure.
 
  • #35
Are those two velocities different? Which one is higher?
 
  • #36
Is the first one higher?
 
  • #37
Why would it be? The motion, as you said, is not accelerated.
 
  • #38
Then they must be equal?
 
  • #39
Indeed. Any part of an unaccelerated path is traversed with the same constant velocity.

Now, you have total horizontal distance and total time.

You also have time to the apex, and and you need to find the horizontal distance to the apex.

Apply the results of the discussion to this.
 
  • #40
So I should set up a proportion of: time it takes to reach max height/x=time of total flight/distance of whole flight
 
  • #41
Exactly.
 
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  • #42
Thanks for helping me.
 

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