Question:

A rock thrown vertically upward from the surface of the moon with initial velocity of 24 m/s reaches a height of \( y(t)=24t-0.8t^2 \)m in \( t \) seconds. What will the maximum height of the rock be?

Solution:

\( v(t)=\frac{dy}{dt}=24-1.6t \)

Set equal to zero to find time of maximum height

\( 24-1.6t=0 \)

\( 1.6t=24 \)

\( t=15s \)

Now plug \( t=15 \) back into our height function to find the maximum height.

\( y(15)=24(15)-0.8(15)^2=360-0.8(225)=360-180=180m \)

Set equal to zero to find time of maximum height

\( 24-1.6t=0 \)

\( 1.6t=24 \)

\( t=15s \)

Now plug \( t=15 \) back into our height function to find the maximum height.

\( y(15)=24(15)-0.8(15)^2=360-0.8(225)=360-180=180m \)