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anonymous
 one year ago
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?
anonymous
 one year ago
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?

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phi
 one year ago
Best ResponseYou've already chosen the best response.0First, 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.

huclogin
 one year ago
Best ResponseYou've already chosen the best response.0解：（1）根据表达式s(t)=16t^264t+80可知，函数s(t)表示球距地面的高度。代入s(t)=0，有： 16t^264t+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))'=32t64 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*164 解得 v0=96ft/sec. ∴落地时球的速度为96ft/sec.
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