Asked by Mithun
A body rises vertically from the earth according to the law s = 64t – 16t2. If it has lost k times its velocity in its 48 ft rise, then k =
Answers
Answered by
Reiny
v = ds/dt = 64 - 32t
at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 16 = 48
so 64k = 48
k = 48/64 = 3/4
at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 16 = 48
so 64k = 48
k = 48/64 = 3/4
Answered by
Mithun
v = ds/dt = 64 - 32t
at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 32= 32
So 64k=32
K= 1/2
at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 32= 32
So 64k=32
K= 1/2
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