Question

A rock is thrown from the side of a ledge. The height (in feet), x seconds after the toss is modeled by: h(x) = -3(x-4)^2+79 How many seconds pass before the rock lands on the ground? Round your answer to the nearest tenths place.

Answer 1

The rock lands on the graound when h(x) = 0:

$h(x)=-3(x-4)^2+79=0$$-3(x-4)^2=-79$

Dividing both sides by -3 gives

$(x-4)^2=\frac{79}{3}$

Taking the square root of both sides gives

$x-4=\sqrt[]{\frac{79}{3}}$

Finally, adding 4 to both sides gives

$x=\sqrt[]{\frac{79}{3}}+4$

converting to a decimal form gives

$x=9.1$

Hence, the rock lands on the ground after 9.1 s.