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Using a Quadratic Equation Problems A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s (t) = 112 + 96t - 16t 2 Complete the table and discuss the interpretation of each point.
t s(t) Interpretation 0 112 s (t) = 112 + 96t - 16t 2 s (t) = 112 + 96(0) – 16(0) 2 s (t) = 112. 0 Sec (t) original height of 112 ft. 0.5 156 s (t) = 112 + 96t - 16t 2 s (t) = 112 + 96(0.5) – 16(0.5) 2 s (t) = 112 + 48 – 4 s(t) = 156. 0.5 sec (t) the ball will reach a height of 156 ft. 1 192 s (t) = 112 + 96t - 16t 2 s (t) = 112 + 96(1) – 16(1)t 2 s (t) = 192. 1 sec (t) the ball is 192 ft. 2 240 s (t) = 112 + 96t - 16t 2 s (t) = 112 + 96(2) – 16(2) 2 s (t) = 240. 2 sec (t) the ball is at 240 ft. -0.12249 100 The ball dropped 12ft and took -0.12249 seconds and proven by two y coordinates See calculations below 6.122 5 100 After thrown the ball went 100ft then to max height and back to down in 6.1225 seconds 1.129 2 200 This represents the throw up taking 1.1292 seconds to 200ft 4.870 8 200 In the arch of the ball going up through 200ft and back down through 200ft, this
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