A ball is thrown upward with an initial velocity of 96 feetper second. The dista

Question

A ball is thrown upward with an initial velocity of 96 feetper second. The distance d (in feet) of the ball from the groundafter t seconds is d(t)=96t-16t^2. (a) When does the ball strike the ground? (b) For what times t is the ball more than 128 feet above theground? A ball is thrown upward with an initial velocity of 96 feetper second. The distance d (in feet) of the ball from the groundafter t seconds is d(t)=96t-16t^2. (a) When does the ball strike the ground? (b) For what times t is the ball more than 128 feet above theground?

Explanation / Answer

the ball strikes the ground when 96t-16t2 =-16t(t-6)=0 so the ball strikes the ground at 6 seconds the ball is more than 128 feet above the ground when 96t-16t2>128 96t-16t2-128>0 -16(t-4)(t-2)>0 between 2 and 4 seconds. I used factoring to find the zeros, but you can also use thequadratic formula

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