JIPMER 2019 Evening Papers Online Practice

(2) Sol. In ascent a = – (g + R)
In decent a = (g – R)
R = acceleration due to air friction
In ascent S = ut1−

1

2

(g + R)t‌12
V = u – (g + R) t‌1 , v = 0, at highest point
u = (g + R) t‌1
2S = (g + R)t‌12 −

1

2

(g+ R)t‌12 ⇒

1

2

(g + R)t‌12
In descent
S =

1

2

(g - R)t‌22 ⇒

1

2

(g + R)t‌12 =

1

2

(g - R)t‌22 ⇒

t1

t2

=√

g−R

g+R

t‌1 < t‌2