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3 years ago

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A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

distance s (in feet) of the ball from the ground after t seconds is 1. After how many seconds does the ball strike the ground?

1 answer:

nordsb [41]3 years ago

7 0

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

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