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THe function h(t)=16t^2+v0t+h0 describes the height in feet above the ground h(t) of an object thrown vertically from a height of h0 feet, with an initial velocity of v0 feet per second, if there is no air friction and t is the

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if an object is projected upward from ground level with an initial velocity of 64 ft per sec,its height h in feet t seconds later is h=16t+64t.after how many seconds is the height 48 ft.

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An object is thrown or fired straight upwards at an initial speed of v_0 ft⁄s will reach height h feet after t seconds, where h and t are related to the formula h=16t^2+v_0 t Suppose the object is fired straight upwards with an

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THe function h(t)=16t^2+v0t+h0 describes the height in feet above the ground h(t) of an object thrown vertically from a height of h0 feet, with an initial velocity of v0 feet per second, if there is no air friction and t is the

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An object is thrown from the top of an 80 foot building with an initial velocity of 64 feet per second. The height (h) of the object after (t) seconds is given by the quadratic equation h=16t+64t+80 .When will the object hit the

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The height of an object projected upward from ground level is given by h=16t squared + 128t. When will the object be 240 feet above the ground? (Please provide steps/explanation to solve this problem.)

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object thrown upward from an initial height of h0 feet with an initial velocity of v0 (in feet per second) is given by the formula h(t)=−16t^2+v0t+h0 feet where t is the amount of time in seconds after the ball was thrown. Also,

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A ball is thrown into the air with an upward velocity of 20 ft/s. Its height (h) in feet after t seconds is given by the function h(t) = –16t^2 + 20t + 2. How long does it take the ball to reach its maximum height? What is the
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