Sign of gravity in free-fall and projectile motion

[clothcut]

Okay, so I just had a quick question regarding sign in two different scenarios. I know what you define as positive or negative is arbitrary, but I wanted to make sure I have signs correct.

Let's say you throw a projectile with initial velocity v at an angle. I assign the velocity as positive, displacement in air as positive and acceeration as negative. What about after the apex when v = 0? Does velocity then become a negative value and g stay the same for in the kinematics equations?

I ask because of this question

"Q: What is the range for a projectile launched at 45 deg angle that has a flight time of 3 seconds?

A. 14.2m
B. 44.1m
C. 72.2m
D. 144.4m

the berkley review "turbo-solution" is
With a flight time of 3 seconds, it takes 1.5 seconds to reach its apex. A 1 second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 second drop covers about 11 to 12m, so the max height is about 11-12m. The range for a 45 deg launch is four times the max height so R ~ 44-48 m."


i see you can use y = 1/2at^2 to obtain 11-12, but why is a positive? Isn't gravity working opposite displacement and thus would be negative? Thanks.

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[BryanNextStep]

If you're going to use the equation you posted, you're solving for the time to fall from the top of the flight back down to the ground. Basically, you're taking:

dy = (Vi)(t) + (1/2)(at^2)

And then we know that at the top of the flight the Vy is equal to zero, so the first term drops out of the equation.

At that point, the ball is falling downwards, so we define "down as positive". That means both d and a can be positive.

If you want to calculate for the upwards part of the flight and define up as positive, then you need to know what Vi is since it doesn't drop out of the equation - at the start of the flight, velocity isn't zero.

 

[Endure]

the berkley review "turbo-solution" is
With a flight time of 3 seconds, it takes 1.5 seconds to reach its apex. A 1 second drop time covers 5m and a 2-i see you can use y = 1/2at^2 to obtain 11-12, but why is a positive? Isn't gravity working opposite displacement and thus would be negative? Thanks.

I found this misleading myself. TBR is using the equation you've posted for the trajectory downward, not upward, causing the negative displacement to cancel with gravity.