The distance d in feet a bomb falls in t seconds is given by d=16t^2/1+0.06t . How many seconds are required for a bomb released at 15,000 feet to reach its target?

The distance d, in feet, a bomb falls in t seconds is given by How many seconds are required for a bomb released at 21,000 feet to reach its target? (If necessary, round your answer to two decimal places.) A) 1167.12 seconds B) 185.76 seconds C) 92.88 seconds D) 2972.19 seconds d=16t^2/(1 +.06t)

21000(1+0.06t)=16t^2 21000+1260t-16t^2=0 4t^2-315t-5250=0 Graphing this I find the zero at 92.88 seconds.

May 14th, 2015

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