| Question |
|---|
| ##t = 3.08s## We have to substitute ##h## by ##0##, as that is the height both when projected as when it hits the ground. Then we solve for ##t##: ##-16t^2 + 48t + 4 = 0##, ##16t^2 – 48t – 4 = 0##, ##4(4t^2 – 12t – 1) = 0##, ##4t^2 – 12t – 1 = 0##. Now we use ##t = (-b +- sqrt((b^2 – 4ac)))/(2a)##: ##t = (- (-12) +- sqrt((-12)^2 – 4(4)(-1)))/(2(4))##, ##t = (12 +- sqrt(144 + 16))/(8)##, ##t = (12 +- sqrt(160))/(8)##, ##t = 3.08s##. Hope it s! 😀 . |
