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

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A baseball is hit at a speed of 30.1 m/s and an angle of 37.2 degrees, with air resistance ignored. The initial vertical and horizontal velocity components are calculated as 18.2 m/s and 24.0 m/s, respectively. The time to reach a height of 10.7 m is determined to be 0.732 seconds and 2.98 seconds, correcting an earlier mistake in the height equation setup. The vertical component of the baseball's velocity at 0.732 seconds is calculated to be 11.03 m/s. The discussion emphasizes the importance of correctly setting up equations and paying attention to signs in calculations.
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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?
 
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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.
 
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