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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?

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  1. phi
    • one year ago
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    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.

  2. huclogin
    • one year ago
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    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.

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