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

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A projectile is fired horizontally from a height of 31.0 m with an initial speed of 310 m/s. It remains in the air for 2.52 seconds and strikes the ground at a horizontal distance of 781 m. The vertical component of its velocity upon impact is calculated to be 24.7 m/s. The confusion arose from mistakenly calculating the resultant velocity instead of focusing solely on the vertical component. The correct answer for the vertical velocity is indeed 24.7 m/s.
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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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