Projectile Motion Question 1: Time, Distance, and Velocity Calculations

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Homework Help Overview

The discussion revolves around a projectile motion problem involving calculations of time, distance, and velocity for a projectile fired horizontally from a height. The original poster presents specific values and seeks assistance particularly with the vertical component of the velocity upon impact.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the vertical component of velocity using equations of motion but expresses difficulty in achieving the correct answer. Some participants clarify that the question specifically asks for the vertical component, while others question whether to use the Pythagorean theorem for combining components.

Discussion Status

Participants are engaged in clarifying the requirements of the problem, with some providing guidance on focusing solely on the vertical component of velocity. There appears to be a productive exchange regarding the interpretation of the question and the calculations involved.

Contextual Notes

The original poster mentions being on their last attempt, indicating a constraint related to the number of tries allowed for the problem. There is also a focus on ensuring the correct interpretation of the question's requirements.

MFlood7356
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1. A projectile is fired horizontally from a gun that is 31.0 m above flat ground, emerging from the gun with a speed of 310 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

I've gotten the correct answers for:
a) 2.52s
b) 781m

I'm really having a problem with part c. I've tried to get the answer three times and I've gotten it wrong every time. I'm on my last attempt and I really need some help.


2. d=-1/2at2, d=vt, v=at, V=sqrtVx2+Vy2, vox =vo cosθ, and voy =vo cosθ

3. Here's my most recent attempt:
v=at v=(9.8)(2.52)= 24.7 V=sqrt3102+24.72=311m/s
 
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Hello M,you have the answer v=at=24.7.The question asked for the vertical component only.
 
So would I find the vertical component of x and y then continue with the Pythagorean theorem? Would the x component be 310m/s?
 
The vertical component of the velocity is 24.7 as you calculated using an equation of motion and that is the answer the question is asking for.

(On hitting the ground the object has a vertical velocity of 24.7 and a horizontal velocity of 310.In your third attempt you used Pythagoras to calculate the resultant velocity but the question didn't ask for this,it asked for the vertical velocity only...24.7m/s)
 
Oh okay thank you it's correct. haha I had it all along
 

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