Asked by sky
A rock is dropped off the edge of a cliff its distance S (in feet) from the top of the cliff after T seconds is S(T) = 16t^2. Assumme the distance from the top of the water below is 96ft.
When will the rock strike the water?
T=16t^2 -96
96=16t^2
80=t^2
4 and squareroot of 5= t
Im not sure if if i did it right because the chapter is about limits
When will the rock strike the water?
T=16t^2 -96
96=16t^2
80=t^2
4 and squareroot of 5= t
Im not sure if if i did it right because the chapter is about limits
Answers
Answered by
bobpursley
You need major help in algebra, not calculus. You may want to drop calculus, if the above work reflects your algebra ability. You made major errors in manipulation and solving that simple equation.
S(t)=16t^2
find t when S=96
96=16^t^2
t^2=96/16 =6
t= sqrt(6)
S(t)=16t^2
find t when S=96
96=16^t^2
t^2=96/16 =6
t= sqrt(6)
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