as a squirrel scurries through a pine tree, it knocks a pine cone loose from a limb 28 meters above the…

as a squirrel scurries through a pine tree, it knocks a pine cone loose from a limb 28 meters above the ground. the pine cone falls to the ground below. to the nearest tenth of a second, how long does it take for the pine cone to hit the ground? hint: use the formula h = -4.9t² + s.

as a squirrel scurries through a pine tree, it knocks a pine cone loose from a limb 28 meters above the ground. the pine cone falls to the ground below. to the nearest tenth of a second, how long does it take for the pine cone to hit the ground? hint: use the formula h = -4.9t² + s.

Answer

Explanation:

Step1: Identify the values

The initial height $s = 28$ meters and when the pine - cone hits the ground $h = 0$. The formula is $h=-4.9t^{2}+s$.

Step2: Substitute values into the formula

Substitute $h = 0$ and $s = 28$ into $h=-4.9t^{2}+s$, we get $0=-4.9t^{2}+28$.

Step3: Rearrange the equation

Rearrange the equation to solve for $t^{2}$: $4.9t^{2}=28$. Then $t^{2}=\frac{28}{4.9}=\frac{280}{49}$.

Step4: Solve for t

Take the square - root of both sides: $t=\sqrt{\frac{280}{49}}$. Since $t>0$ (time cannot be negative in this context), $t=\frac{\sqrt{280}}{7}\approx\frac{16.733}{7}\approx2.4$.

Answer:

$2.4$