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) = -16t² + 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 derivative definition

The limit definition of the derivative is \( h'(t)=\lim_{h \to 0}\frac{h(t + h)-h(t)}{h} \). First, find \( h(t + h) \) for \( h(t)=-16t^{2}+50t + 5 \).

\( h(t + h)=-16(t + h)^{2}+50(t + h)+5=-16(t^{2}+2th+h^{2})+50t + 50h+5=-16t^{2}-32th-16h^{2}+50t + 50h+5 \)

Step2: Compute \( h(t + h)-h(t) \)

Subtract \( h(t) \) from \( h(t + h) \):

\( h(t + h)-h(t)=(-16t^{2}-32th-16h^{2}+50t + 50h+5)-(-16t^{2}+50t + 5)=-32th-16h^{2}+50h \)

Step3: Divide by \( h \) ( \( h

eq0 \))
\( \frac{h(t + h)-h(t)}{h}=\frac{-32th-16h^{2}+50h}{h}=-32t-16h + 50 \)

Step4: Take the limit as \( h\to0 \)

\( h'(t)=\lim_{h \to 0}(-32t-16h + 50)=-32t + 50 \)

Step5: Evaluate at \( t = 1 \)

Substitute \( t = 1 \) into \( h'(t) \): \( h'(1)=-32(1)+50 = 18 \)

The derivative \( h'(t) \) represents the velocity of the softball. At \( t = 1 \), \( h'(1)=18 \) ft/sec, which is positive. A positive velocity (in the context of upward - initial motion) means the object is still moving upward.

Answer:

\( h'(1)=18 \)

Part B