A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is
Let the initial velocity of ball be u
Time of rise t1=ug+a and height reached =u22(g+a)
Time of fall t2 is given by
12(g−a)t22=u22(g+a)
⇒ t2=u√(g+a)(g−a)=u(g+a)√g+ag−a
∴ t2>t1 because 1g+a<1g−a