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I don't understand this at all! use the position equation s=16t^2 +v0t+s0 to write a function that represents the situation. Find the average rate of change from t1 to t2. a) An object is thrown upward from a height of 6 feet at
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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)=âˆ’16t^2+v0t+h0 feet where t is the amount of time in seconds after the ball was thrown. Also,
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In physics, one learns that the height of an 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)=âˆ’16t^2+v0t+h0 feet where t is the amount of
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somebody please help me find the solution totally lost.... An object of mass 100g is thrown vertically upward from a point 60cm above the earths surface with an initial velocity of 150cm/sec. It rises briefly and then falls
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if an object is thrown vertically upward with an initial velocity of v from an original position of s, the height h at any time t is given by: h=16t^2+vt+s (where h and s are in ft t is in seconds and v is in ft/sec) If a rock is
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