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An object is projected directly upward from the ground. After t seconds it’s distance in feet above the ground is s=144t16t^2 How long does it take the object to reach its maximum height? What is the maximum height the object
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An object is projected vertically upward from the top of a building with an initial velocity of 112 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the equation s(t) = −16t^2 + 112t + 100. (a) Find
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An object is projected vertically upward from the top of a building with an initial velocity of 144 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the function s(t) = −16t2 + 144t + 120. A) Find
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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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A ball is projected upward from the ground. Its distance in feet from the ground in t seconds is given by s(t) = 16t^(2)+128t At what will the ball be at 213ft from the ground?
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The formula h=vt−16t^2 gives the height h, in feet, of an object projected into the air with an initial vertical velocity v, in feet per second, after t seconds. If an object is projected upward with an initial velocity of 112
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Please help me solve this: If an object is projected upward with an initial velocity of 112 feet per second from a height h of 128 feet, then its height t seconds after it is projected is defined by the equation h=16t^2 + 112t +
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Please help if you can. If a ball is shot projected straight upward at a rate of 64 feet per second from a rooftop, 80 feet above ground level, the height of feet above ground is modeled by the equation, h(t)=16t^2+64t+80 Find
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