Projectile Questions: Solving Max Height & Xmax

  • Thread starter KingPapaya
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In summary, the equation for maximum height is Delta y=VyoT-.5gt^2, where Vyo is the initial vertical velocity and T is the time it takes to reach the maximum height.
  • #1
KingPapaya
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Homework Statement



We did an experiment where we threw a tennis ball far and high.
For the far toss:
initial height=1.3m
time it took to land=1.92sec
distance traveled=23.4m

For the high toss:
initial height=1.37m
time it took to land=2.61sec
distance traveled=9.5m

Now I have to calculate the maximum height for both tosses.
Then I have to solve for the xmax at maximum height for the long toss.

Homework Equations


I know that I have to use delta y=VyoT-.5gt^2 for maximum height.


The Attempt at a Solution


I know the formula to solve for maximum height but there isn't any angle that can be used so I don't know what to do. I tried setting Vyo equals to 0 but that gave me a negative number. For the xmax, I know it is supposed to be less than half of the distance traveled because of the initial height but I don't know how to find it.
 
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  • #2
If you only care about the max height, ignore the horizontal distance part. Focus solely on the vertical motion. The equation you have is good. What is known and unknown in it?
 
  • #3
Hello, KingPapaya. Welcome to PF!

Consider the far toss. You know the equation that relates y and t. Apply that equation for t = 1.92 s. What can you deduce from that equation?

Repeat for the equation that relates x and t.

[Oops, I see voko already replied while I was constructing my reply.]
 
  • #4
Delta y is y-yo so I have yo, which is 1.3m for far/long and 1.37m for high.
I have t, which is 1 .92 sec for far/long and 2.61 sec for high.
G is 9.8m/s^2.
I'm supposed to be solving for y but I'm not sure what Vyo is
 
  • #5
Since the ball hits the ground, do you think you don't have ## \Delta y ##?
 
  • #6
I thought I have Δy because I have yo
 
  • #7
Thus you have everything in that equation - except ## v_{0y} ##. Can you find ## v_{0y} ## then?
 
  • #8
So Vyo=(Δy+.5gt^2)/t?

Vyo=(y-1.3m+.5(9.8ms^2)(1.92sec))/1.92sec

Wait so what do I do about y, since it is also unknown
 
  • #9
Where is the ball at t = 1.92 s?
 
  • #10
Oh so since it is on the ground, Δy is -1.3m because 0-1.3m?
 
  • #11
Correct!
 
  • #12
After plugging in the numbers to the equation the number I get for long and high are the maximum heights?
Vyo=maximum height?
 
  • #13
## v_{0y} ## is the initial vertical velocity.

What is the vertical velocity at the max height?
 
  • #14
Would it be 0?
 
  • #15
That is correct as well. Now that you know the initial and the final velocities, and you know the acceleration as well, can you find the time?
 
  • #16
0=Vyo-gt?
So T is (0-Vyo)/-g?
 
  • #17
Yes.
 
  • #18
Now what should I do with the time and Vyo solved for to get the maximum height?
Will I have to use the original equation?
 
  • #19
Is there any reason you cannot?
 
  • #20
No.
So after I do the same thing for the high toss, should I make a triangle?
 
Last edited:
  • #21
I am not sure what kind of triangle you have in mind.
 
  • #22
Is xmax the same as the range?
 
Last edited:
  • #23
Let's recap on the data you have that are relevant for the horizontal motion: total time of flight; time to the apex; total horizontal distance. You need to find the horizontal position of the apex.

Any ideas? What kind of motion is the horizontal motion?
 
  • #24
I'm not sure
 
  • #25
Is the horizontal motion accelerated?
 
  • #26
The horizontal motion isn't accelerated
 
  • #27
Then if the ball covers some distance in time ##t_1## and some other distance in time ##t_2##, how are the distances related?
 
  • #28
If you add the distances together, you get the total distance traveled?
 
  • #29
That does not relate the distances. Say the first distance is ## s_1 ## and the second is ## s_2 ##. What is ## \frac {s_1} {t_1} ##? What is ## \frac {s_2} {t_2} ##?
 
  • #30
s1/t1 and s2/t2 both equal speed?
 
  • #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} ##?
 
  • #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?
 

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