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 distances(in feet) of the ball from the ground aftertseconds is s (t)= 112 + 96t - 16t2 Complete the table and discuss the interpretation of each point.

ts(t)Interpretation 0112 s (t)= 112 + 96t - 16t2 s (t)= 112 + 96(0) – 16(0)2 s (t) = 112.0 Sec (t) original height of 112 ft. 0.5156 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. 1192 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. 2240 s (t)= 112 + 96t - 16t2 s (t)= 112 + 96(2) – 16(2)2 s (t) = 240. 2 sec (t) the ball is at 240 ft. -0.12249100The ball dropped 12ft and took -0.12249 seconds and proven by two y coordinates See calculations below 6.1225100After thrown the ball went 100ft then to max height and back to down in 6.1225 seconds 1.1292 200This represents the throw up taking 1.1292 seconds to 200ft 4.8708200In the arch of the ball going up through 200ft and back down through 200ft, this

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