Sat June 18, 2011 By: Mayank Gupta

A stone is thrown up with an initial speed v0. There is a resisting acceleration k(v^1/2) due to air, where v is instantaneous velocity and k is some positive constant. Find the time taken to reach the highest point and the maximum height attained by the particle.(neglect acceleration due to gravity)

Expert Reply
Tue June 21, 2011

Using first equation of motion

v’ = u - a t

Where v’ is the speed at highest point and t is the time to reach that point. As acceleration due to gravity is neglected, then  

0 = v0 - k(v^1/2) t

t = v0 / k(v^1/2)

Let H is the maximum height attained by the stone then,

2 = u2 – 2aH

0 = (v0)2 – 2 k(v^1/2) H

H = (v0)2 / 2 k(v^1/2)

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