A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. Its height h, in feet, after t seconds is given by h(t)=−16t^2 +96t +640. After how long will the ball reach the ground?

check the picture below.

thus is at 0 = -16t² + 96t +640,

[tex]\bf 0=-16t^2+96t+640\implies 0=-16(t^2-6t-40) \\\\\\ 0=t^2-6t-40\implies 0=(t-10)(t+4)\implies t= \begin{cases} \boxed{10}\\ -4 \end{cases}[/tex]

well, clearly it can't be a negative value for the elapsed seconds, so it can't be -4.

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A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. Its height​ h, in​ feet, after t seconds is given by ​h(t)=−16t^2 +96t +640. After how long will the ball reach the​ ground?

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A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. Its height h, in feet, after t seconds is given by h(t)=−16t^2 +96t +640. After how long will the ball reach the ground?