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)
Asked by Mayank Gupta | 18th Jun, 2011, 02:46: AM
Expert Answer:
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,
v2 = u2 2aH
0 = (v0)2 2 k(v^1/2) H
H = (v0)2 / 2 k(v^1/2)
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,
v2 = u2 2aH
0 = (v0)2 2 k(v^1/2) H
H = (v0)2 / 2 k(v^1/2)
Answered by | 21st Jun, 2011, 09:23: AM
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