# Module_4_Group_Disscussion.docx.docx - Using a Quadratic...

• 5

This preview shows page 1 - 3 out of 5 pages.

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 1.) s (t) = 112 + 96t - 16t 2 2 112 112 96 16 t t 96 16 0 t t t (96 16 ) 0 t t 96 16 t 6 t 2.) t = 0.5 s (t) = 112 + 96t - 16t 2 2 112 96(0.5) 16(0.5) s t 2 112 96(0.5) 16(0.5) s t 112 48 4 s t 156 s t 3.) t = 1 s (t) = 112 + 96t - 16t 2 2 112 96(1) 16(1) s t 2 112 96(1) 16(1) s t 112 96 16 s t 192 s t 4.) t = 2 s (t) = 112 + 96t - 16t 2 2 112 96(2) 16(2) s t 2 112 96(2) 16(2) s t 112 192 64 s t s (t) = 240 ft Step 1.) Replace s(t) with 112 ft so we can solve for time (t). Step 2.) Subtract 112 from both sides of the equation allowing us t use the zero-product principle. Step 3.) Factor out a (t) Step 4.) t = 0 meaning at 0 seconds the ball is at its original height of 112ft Step 5.) Divide both sides by 16 Step 6.) At 6 seconds the ball has returned to its original height of 112 ft Step 1.) Replace (t) with 0.5 ft so we can solve for distance s(t). Step 2.) Multiply 96 and by 0.5 and -16 by 2 (0.5) Step 3.) Add and subtract the remaining numbers Step 4.) s(t) = 156 meaning at 005 seconds the ball is at a height of 156 ft Step 1.) Replace (t) with 1 ft so we can solve for distance s(t). Step 2.) Multiply 96 and by 1 and -16 by 2 (1) Step 3.) Add and subtract the remaining numbers Step 4.) s(t) = 192 meaning at 1 seconds the ball is at a height of 192 ft Step 1.) Replace (t) with 2 ft so we can solve for distance s(t). Step 2.) Multiply 96 and by 2 and -16 by 2 (2) Step 3.) Add and subtract the remaining numbers Step 4.) s(t) = 240 meaning at 2 seconds the ball is at a height of 240 ft
5 & 6.) s(t) = 100 s (t) = 112 + 96t - 16t 2 2 100 112 96 16 t t 2 100 112 96 16 0 t t 2 12 96 16 0 t t 2 4 24 12 0 t t 2 ( 24) ( 24) 4 4( 3) 2 4 t   2 24 ( 24) 4 4( 3) 2 4 t