Vertical resisted motion question help (1 Viewer)

[Lith_30]

How do I do part iv?

 

[notme123]

you can simplify the expression in b (iii) by letting to resemble f(x)
so 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

 

[Lith_30]

you can simplify the expression in b (iii) by letting

to resemble f(x)
so

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

I think I get it

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?

 

[notme123]

I think I get it

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