How Does Kinematics Calculate the Trajectory of a Thrown Ball?

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Homework Help Overview

The discussion revolves around a kinematics problem involving the trajectory of a ball thrown upward with an initial velocity. Participants are analyzing calculations related to the time to reach maximum height, maximum height, and the velocity upon returning to the ground.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants present various calculations for the time to maximum height, maximum height, and return velocity. There are discussions about the use of significant figures and rounding in calculations, as well as the correct formulas to apply for constant acceleration.

Discussion Status

There is an ongoing examination of the calculations presented, with some participants questioning the accuracy of the formulas used and the rounding methods. Suggestions for clearer, more precise calculations are offered, and there is a recognition of differing instructional approaches regarding rounding.

Contextual Notes

Participants mention specific instructions regarding rounding and the use of unrounded values in calculations, indicating a potential source of confusion. There are also references to grading practices that may affect how answers are evaluated.

tiarakiyoshi
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Please tell me if I have done this correctly.

A ball is thrown upward from the ground with an initial velocity of 10 m/s. Assume all digits are significant.

a. How long before the ball reaches its maximum height?

t = (0m/s - 10m/s) / -9.80m/s2 t = 1.0s

b. What is the ball's maximum height?

x = 0m + .5(10m/s + 0m/s) x = 5.0 m

c. What is the velocity of the ball when it returns to the ground?

v = 0m/s + -9.80m/s2 * 1.0s v = -9.8 m/s
 
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What have you done in b?

Never use rounded values (in the calculator), just give them as answers.
b and c are wrong due to earlier roundings.
 
We were instructed to round after each section of the problem. In b I used the formula x = x0 + 1/2(v0 + v)t. x0 is x naught and so on.
 
You are supposed to round the answer, but I've always been instucted to use unrounded values in the calculator.

Anyways, the formula you used is wrong.

The correct one (works for constant acceleration):
[tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]
 
x = 0m/s + (10m/s * 1.0 s) + (-9.80m/s2 * 1.0s^2)/2 = 5.1 m

Ok, is this right?
 
Almost correct, the unit of x0 is not m/s, but m (you have x = 0m/s ...).
It doesn't effect the result though, so: Yes, it's correct.
 
It is better to write clean, nice and exact calculations.
In some places they take off points in an exam if you don't do so .
It aslo better to slove the question step by step and in a general form and insert numbers only at the final stage.
Here is how I would slove this question:
a.
0) [tex]V_0=10m/s[/tex]

[tex]V=0m/s[/tex]

[tex]a=-9.8m/s^2[/tex]

1) [tex]V=V_0+at[/tex]

2) [tex]0=10-9.8t[/tex]

3) [tex]10=9.8t[/tex]

4) [tex]t=\frac{10}{9.8}=1.02...s[/tex]

b.

1) [tex]X=X_0+V_0t+\frac{1}{2}at^2[/tex]

2) [tex]X=0+10*1.02+\frac{1}{2}*-9.8*1.02^2[/tex]

3) [tex]X=12-5.09[/tex]

4) [tex]X=6.91m[/tex]

c. I think it is not hard find out (even without calculations)that the answer is -10m/s, try to figure out why by yourself (it is really easy).

A little tip: Always make a drawing.
 
Last edited:
What did you, Alkhimey, do in b 3) ?
 
Thank you everyone for your help. I know it's a simple problem. My instructor only gave me 2 points out of 20 for the answer, and I'm fairly certain that isn't reasonable. I've suspected that he was incompetent, but that's a different matter. I decided it would be best to take the course at a later date with a better instructor.
 
  • #10
Päällikkö said:
What did you, Alkhimey, do in b 3) ?
[tex]10*1.02=12[/tex]
[tex]1/2*-9.8*1.02^2=5.09[/tex]
Therefore:
[tex]X=12-5.09[/tex]
 

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