A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is.

Options

A. Equal to the time of fall
B. Less than the time of fall
C. Greater than the time of fall
D. Twice the time of fall

Answer

B

Explanation

Let initial velocity = u
time to rise `t_1 = (u)/((g+a))`
maximum height reached
`h = (u^2)/(2(g+a))`
For time to all
`1/2 (g - a)t_2^2 = (u^2)/(2(g + a))`
`rArr t_2^2 = (2u^2)/(2(g^2 - a^2)) rArr t_2 (u)/(sqrt(g^2 - a^2))`
`t_2 = (u)/((g+a)) sqrt((g+a)/(g-a)) rArr t_2 = t_1 sqrt((g+a)/(g-a))`
`therefore t_2 gt t_1`