Solving Projectile Questions: How Long in Air & Distance to Ground (c)"

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SUMMARY

The discussion focuses on solving projectile motion problems, specifically a projectile fired horizontally from a height of 43.0 m with an initial speed of 300 m/s. The participant successfully calculated the time in the air as 3 seconds and the horizontal distance traveled as 890 m. However, they encountered difficulties in determining the magnitude of the vertical component of the velocity upon impact with the ground. The correct approach involves using the formula vy = v0y + ayt, where v0y is zero for horizontal launch, leading to a vertical velocity of approximately -29.4 m/s at impact.

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Homework Statement


A projectile is fired horizontally from a gun that is 43.0 m above flat ground, emerging from the gun with a speed of 300 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?


Homework Equations


I already solved both (a) and (b) by using the vy=v0y + ayt and vx=vox + axt, i got 3 seconds for (a) and 890m for (b), I just don't know how to solve for (c)


The Attempt at a Solution


For (c), I tried using the vy=v0y + ayt formula and plugging in the time as 3 seconds and i got an answer of -29.4 which came out to be wrong, i also plugged in 300m/s as the v and multiplied it by sin0. Let me know what I'm doing wrong on this problem, I don't know any other way to look at it.
 
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shell4987 said:
For (c), I tried using the vy=v0y + ayt formula and plugging in the time as 3 seconds and i got an answer of -29.4 which came out to be wrong
They just want the magnitude.
 

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