Projectile Motion: Solving for Time & Velocity with No Air Resistance

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a baseball hit at a specific speed and angle, where participants are tasked with finding the times the baseball reaches a certain height and calculating its vertical velocity at an earlier time. The subject area is kinematics, specifically focusing on motion under constant acceleration without air resistance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of equations to find the time at which the baseball reaches a height of 10.7 m, with some questioning the initial conditions used in the calculations. There are attempts to clarify the use of the quadratic formula and the interpretation of negative time values. Others suggest revising the approach to finding the vertical component of velocity.

Discussion Status

There is an ongoing exploration of different methods to solve for time and velocity, with some participants providing alternative equations and corrections to previous attempts. While some new times have been proposed, there is no explicit consensus on the correctness of the calculations, and participants continue to engage with the problem.

Contextual Notes

Participants note that this is part of a larger multi-part question, indicating that further assistance may be needed as the discussion progresses. There are also mentions of potential misunderstandings regarding the setup of the equations and the interpretation of results.

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Homework Statement



A major leaguer hits a baseball so that it leaves the bat at a speed of 30.1 m/s and at an angle of 37.2 degrees above the horizontal. You can ignore air resistance.

a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?

Voy = 18.2 m/s (30.1 sin37.2)
Vox= 24.0 m/s (30.1 cos37.2)

ax=0
ay= -9.8 m/s^2

yo = 0
xo = 0

vy= -18.2 m/s
xy= 24.0 m/s

Homework Equations



any constant acceleration equations.

The Attempt at a Solution



I received Ttotal = 3.71 seconds which I think is right.

I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2

which is -4.9t^2 + 18.2t + 10.7

I did the quadratic equation but received the times -0.516 and 4.23, and added together equal my total time of 3.71, however, I don't understand how there is a negative number.

Are these the right times?

P.S (this is a 7 part question, so I might need help after this question).

Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

I'm guessing this means use .732 seconds for time.
I used the equation: vy=voy +ayt

Vy= 18.2+(-9.8)(.732)

=11.03 m/s.

Is this right?
 
Last edited:
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Hey!
I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2
Can you please explain how you arrived at substracting your desired height from zero?
It should be, just 10.7, nothing else.
Daniel
 
Remember: Vy = Voy - gt

I think you should revise your Vy => Although for this problem it is not necessary

use this
Yf = Yo + Voyt - .5gt^2

just think where you want your Yf to be and solve for t
 
danielakkerma said:
Hey!

Can you please explain how you arrived at substracting your desired height from zero?
It should be, just 10.7, nothing else.
Daniel

I received the new times of 0.732 and 2.98 which sounds much better!

I should have had 10.7-0, not 0-10.7.
 
cd.riter said:
Remember: Vy = Voy - gt

I think you should revise your Vy => Although for this problem it is not necessary

use this
Yf = Yo + Voyt - .5gt^2

just think where you want your Yf to be and solve for t

sounds much easier than my method! thanks.
 
Always look for that blasted minus sign!
Well done!
Daniel
 
danielakkerma said:
Always look for that blasted minus sign!
Well done!
Daniel

thanks!
 
Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

I'm guessing this means use .732 seconds for time.
I used the equation: vy=voy +ayt

Vy= 18.2+(-9.8)(.732)

=11.03 m/s.

Nevermind, I know this is right.
 
Last edited:

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