A ball is thrown vertically upward from the top of a 100-foot tower, with an initial velocity of 20 ft/sec. Its position function is s(t) = –16t2 + 20t + 100. What is its velocity in ft/sec when t = 1 second? (This is Calculus, involving limits, please help and explain, because I mostly just need to know how to do this. :) )



Answer :

Answer: -12 ft/sec

Definition of a limit

f'(x) = [tex]\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

In this problem, velocity is s'(t): the derivative of the position function.

s'(t) = [tex]\lim_{h \to 0} \dfrac{s(t+h)-s(t)}{h}[/tex]

v(t) = [tex]\lim_{h \to 0} \dfrac{s(t+h)-s(t)}{h}[/tex]

Substituting the equation

s(t+h) = -16(t+h)² + 20(t+h) + 100

s(t) = -16t² + 20t + 100

v(t) = [tex]\lim_{h \to 0} \dfrac{[-16(t+h)^{2}+20(t+h)+100]-[-16t^{2}+20t+100]}{h}[/tex]

Simplify

[tex]\lim_{h \to 0} \dfrac{[-16t^{2}-32th-16h^{2}+20t+20h+100]-[-16t^{2}+20t+100]}{h}[/tex]

= [tex]\lim_{h \to 0} \dfrac{-16t^{2}-32th-16h^{2}+20t+20h+100+16t^{2}-20t-100}{h}[/tex]

Several terms cancel out:

= [tex]\lim_{h \to 0} \dfrac{-32th-16h^{2}+20h}{h}[/tex]

Factor out an h from the numerator:

= [tex]\lim_{h \to 0} \dfrac{(-32t-16h+20)h}{h}[/tex]

The h's cancel out:

= [tex]\lim_{h \to 0} \(-32t-16h+20[/tex]

Find the limit. All terms with an h will equal to 0.

= -32t + 20

v(t) = -32t + 20

Find v(1)

To find the velocity when t = 1, we need to evaluate v(1):

v(1) = -32(1) + 20

= -32 + 20

= -12 ft/sec