1 If an object is launched straight up into the air from a s

1. If an object is launched straight up into the air from a starting height of h0 feet, the the height of the obkect after t seconds is approximately h(t)=-16t^2+v0t + h0 feet, where v0 is the intial velocity of the object. Find the starting height and intial velocity of an object that attains a maximum height of 562 feet four seconds after being launched.

2. The concentration of a certain drug in the bloodstream t hours after the durg is administered is given by c(t)=35gte^(-bt) ng/ml, where b is some positve constant. Suppose that the drug reaches maximum concentration exactly seven hours after being administered. find the constant b.

Solution

a) h(t)= -16t² +v₀t+h₀

given that-> at 4 seconds h=562 feet

562= -16*4*4+v₀*4+h₀

4v₀+h₀=818 ----------equation 1

Further, d( h(t))/dt = -32t+v₀

at maxima -32t+v₀ =0 ; i.e. v₀= 32t= 32*4= 128 feet/sec

substituting in equation 1 ->> 4*128+h₀ = 818 =>> h₀ =818-4*128 = 306 feet

b) c(t) = 35gte^(-bt)

d(c(t))/dt = 35ge^(-bt)+-35bgte^(-bt)

at maxima; d(c(t))/dt = 0 =>> 35ge^(-bt) - 35bgte^(-bt)

= 35ge^(-b7) - 35bg7e^(-b7)

=>1=b7

=>b =1/7