Question

1. when throwing a softball directly upward from a height of 5 ft with an initial velocity of 50 ft/sec, the height of the softball after t seconds is given by ( h(t) = -16t^2 + 50t + 5 ) (until the ball hits the ground).

a. using the limit definition of the derivative, find ( h(1) ).
b. using your answer in part a, is the ball going up or going down one second after being thrown?

Explanation:

Part A

Step1: Recall limit definition of derivative

The limit definition of the derivative of a function \( h(t) \) at a point \( t = a \) is \( h'(a)=\lim_{h
ightarrow0}\frac{h(a + h)-h(a)}{h} \). Here, \( a = 1 \), so we need to find \( \lim_{h
ightarrow0}\frac{h(1 + h)-h(1)}{h} \).

First, find \( h(1+h) \) and \( h(1) \).

Given \( h(t)=- 16t^{2}+50t + 5 \).

For \( h(1 + h) \):
Substitute \( t=1 + h \) into \( h(t) \):
\( h(1 + h)=-16(1 + h)^{2}+50(1 + h)+5 \)
Expand \( (1 + h)^{2}=1 + 2h+h^{2} \):
\( h(1 + h)=-16(1 + 2h+h^{2})+50 + 50h+5 \)
\(=-16-32h-16h^{2}+50 + 50h+5 \)
Combine like terms:
\( (-16 + 50+5)+(-32h + 50h)-16h^{2} \)
\( =39 + 18h-16h^{2} \)

For \( h(1) \):
Substitute \( t = 1 \) into \( h(t) \):
\( h(1)=-16(1)^{2}+50(1)+5=-16 + 50+5 = 39 \)

Step2: Substitute into the limit formula

Now, \( \frac{h(1 + h)-h(1)}{h}=\frac{(39 + 18h-16h^{2})-39}{h} \)
Simplify the numerator:
\( \frac{39 + 18h-16h^{2}-39}{h}=\frac{18h-16h^{2}}{h} \)
Factor out \( h \) from the numerator (assuming \( h
eq0 \), which is valid for the limit as \( h
ightarrow0 \) but \( h
eq0 \) in the fraction):
\( \frac{h(18 - 16h)}{h}=18-16h \) (for \( h
eq0 \))

Step3: Evaluate the limit

Now, find \( \lim_{h
ightarrow0}(18-16h) \)
As \( h
ightarrow0 \), \( - 16h
ightarrow0 \), so the limit is \( 18-0 = 18 \)

The derivative \( h'(t) \) represents the velocity of the softball at time \( t \). If \( h'(t)>0 \), the object is moving upward (since the height function's derivative is velocity, positive velocity means upward motion, negative means downward).

We found in Part A that \( h'(1) = 18 \), and \( 18>0 \).

Answer:

\( h'(1)=18 \)

Part B