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Vertical resisted motion question help (1 Viewer)
- Thread starter Lith_30
- Start date
you can simplify the expression in b (iii) by letting 
to resemble f(x)
so >0)
a (iii)
therefore
so 
when 
, or in other words, its still above ground.
disregarding the first T seconds, this means the time it took to reach max height (T seconds), is not long enough for the object to reach the ground. so that means it takes longer to fall to the ground than rise.
theres probably a better explanation
so
therefore
disregarding the first T seconds, this means the time it took to reach max height (T seconds), is not long enough for the object to reach the ground. so that means it takes longer to fall to the ground than rise.
theres probably a better explanation
I think I get ityou can simplify the expression in b (iii) by lettingto resemble f(x)
soa (iii)
thereforeso
when
, or in other words, its still above ground.
disregarding the first T seconds, this means the time it took to reach max height (T seconds), is not long enough for the object to reach the ground. so that means it takes longer to fall to the ground than rise.
theres probably a better explanation
Since y>0, this means that the body is still above the ground despite 2T seconds passing. Since it took T seconds to go up that means the other T seconds were spent going down but it wasn't able to cover the same distance as going up. Therefore the downward journey takes longer.
Is this explanation right?
ye thats right