Projectile is fired over level ground

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A projectile is launched with a vertical velocity of 20 m/s and a horizontal velocity of 30 m/s, resulting in a total flight time of 4 seconds. The calculations confirm that the horizontal displacement is 120 meters, using the equation for motion. The method applied to determine time and distance appears correct, with no identified errors in the analysis. The discussion highlights the importance of accurately applying kinematic equations in projectile motion problems. Overall, the solution is validated and deemed satisfactory.
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Homework Statement


A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s2, the distance from launching to landing points is:

Homework Equations


Vi(t) +0.5at^2

The Attempt at a Solution


I used the first equation to solve for the time it took from start to finish. 0=20m/s(t)+0.5(-10m/s2)t^2 and that gave me 4 seconds to launch and land.

Next i used the same equation only now i used the x component so x=30m/s(4s) and the displacement was 120 meters.

Any errors you see?
 
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I'm guessing you made a typo in your relevant equations...
 
Joppy said:
I'm guessing you made a typo in your relevant equations...
What do you mean?
 
Le_Anthony said:

Homework Statement


A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s2, the distance from launching to landing points is:

Homework Equations


Vi(t) +0.5at^2

The Attempt at a Solution


I used the first equation to solve for the time it took from start to finish. 0=20m/s(t)+0.5(-10m/s2)t^2 and that gave me 4 seconds to launch and land.

Next i used the same equation only now i used the x component so x=30m/s(4s) and the displacement was 120 meters.

Any errors you see?
Looks good to me !

I like the analysis.
 
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SammyS said:
Looks good to me !

I like the analysis.
Thanks!
 
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