dY = vi*t + 1/2at^2
dY = 6m/s*0.612s + 1/2(-9.8m/s^2)(0.612s^2)
dY = 3.672m + (-1.83m) = 1.842m
So my answer is 1.842?
Yes, it should be 1.842 meters
I took the 0.612 seconds from the time that I calculated in your previous question
Here vi is your initial velocity. t is the time that was claculated before. and a is acceleration
Does that make sense?
I wanted to inform you, if you don't already know, that Bilal is using the wrong equations in his answers. For example on the throwing the ball question, it is impossible that a ball could travel 25 meters and not have any initial velocity.
That I was just using yours and the Micheals guys answers
Okay, good! Just wanted to make sure you weren't using the wrong thing. :)
Why did the ball experience no acelleration?
You're welcome! :)
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