Question

The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = -16t2 - 64t + 80, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?

Answer 1

First, find the time "t" when the ball hits the ground (i.e. when s(t)= 0) by solving
for "t" in
\[ -16t^2 -64t+80- 0 \]

Next, find the derivative of the height with respect to t, \( \frac{d s}{dt} \) This is the instantaneous velocity of the ball as a function of t.

Evaluate this expression at the time found from the first step.

Answer 2

解：（1）根据表达式s(t)=-16t^2-64t+80可知，函数s(t)表示球距地面的高度。代入s(t)=0，有：
-16t^2-64t+80=0
解得 t1=1， t2=-5（舍去）.
∴球在空中运动的时间为1s.
（2）∵s(t)是位移关于时间的函数，∴对时间t求导，得到瞬时速度关于时间t的函数，即：
v=(s(t))'=-32t-64
由（1）得，球在空中运动的时间为t=1s，即第1s末球落地.代入t=1s,有：
v0=-32*1-64
解得 v0=-96ft/sec.
∴落地时球的速度为-96ft/sec.