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

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A projectile is fired horizontally from a height of 43.0 m with an initial speed of 300 m/s. It remains in the air for 3 seconds and strikes the ground at a horizontal distance of 890 m. To find the vertical component of its velocity upon impact, the formula vy = v0y + ayt is used, but the initial vertical velocity (v0y) is zero since the projectile is fired horizontally. The correct calculation for the vertical component results in a magnitude of approximately 29.4 m/s downward, indicating a misunderstanding in the sign convention. The key takeaway is that only the magnitude is required for the final answer.
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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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