A basketball is thrown with an initial upward velocity of 23 feet per second from a height of 7 feet above the ground. The equation h=−16t2+23t+7h=−16t2+23t+7 models the height in feet t seconds after the basketball is thrown. After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground. About how long after it was thrown does it go into the hoop? Select one: a. 1.29 seconds b. 1.44 seconds c. 1.70 seconds

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10=-16t^2+23t+7  subtracting the right side from both sides

16t^2-23t+3=0  using the quadratic formula

t=(23±√337)/32, since we are looking for the larger value of t, as the ball in returning downward...

t=(23+√337)/32

t≈1.29 seconds