At height 48, we have
64t-16t^2 = 48
t=1 or 3
Since the body is rising (not falling), t=1
v(1) = 64-32*1 = 32
32 = 1/2 * 64
so, k = 1/2
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 =
A)1 B) 1/2 C) 1/3 D)1/9
2 answers
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