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more information is required to answer this question like her initial speed(or velocity),acceleration,frictional force(if any) etc
No it isn't. Assuming no air resistance, you just use: \[v^2=u^2+2as\]Her initial vertical velocity is zero so this term disappears, you know s=10m and a=9.81ms^-2
the woman is driving off the platform..how can her inital velocity be zero?
I agree with bendt ans its correct
Her initial vertical velocity is zero, not her horizontal velocity.
before driving off the plateform her speed is zeroo
At the instant she leaves the platform, the vertical component of her velocity vector is zero.
mmm...but isnt the final speed affected by both vertical and horizontal components?
horizontal component doesnot play any role in vertical velocity
Yes, Vids93, but that's not what the question's asking. Think of her walking off the platform.
i see the qs says what is the 'speed' as the woman hits the water!
Yes but it clearly doesn't want you to take into account the horizontal velocity because, as you said, you would need more information.
What would the u^2 stand for?
Ok thank you!