A stone is dropped from the upper observation deck of a tower, 800 m above the ground. (Assume g = 9.8 m/s2.)

Question

(d) If the stone is thrown downward with a speed of 6 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

Answer 1

The acceleration of the stone, a, is a constant function of time given by:

a(t) = -9.8

The velocity of the stone is given by:

v(t) = ∫ a(t) dt = ∫ -9.8 dt = -9.8t + C

"the stone is thrown downward with a speed of 6 m/s"

so v(0) = -6

⇒ -9.8 * 0 + C = -6

⇒ C = 6

so the velocity of the stone is:

v(t) = -9.8t - 6

The position of the stone is given by:

s(t) = ∫ v(t) dt = ∫ (-9.8t - 6) dt = -4.9t² - 6t + C

"A stone is dropped from the upper observation deck of a tower, 800 m above the ground."

so s(0) = 800

⇒ -4.9 * 0² - 6 * 0 + C = 800

⇒ C = 800

so the position of the stone is:

s(t) = -4.9t² - 6t + 800

We want to know how long it takes the stone to reach the ground;

that is, at what time t is the position of the stone 0?

So we solve the equation s(t) = 0.

s(t) = 0

⇒ -4.9t² - 6t + 800 = 0

⇒ t = {-(-6) ± [(-6)^2 - 4 * (-4.9) * 800]^0.5 } / [2 * (-4.9)] using the quadratic formula

= [6 ± (36 + 15,680)^0.5] / (-9.8)

= (6 ± 15,716^0.5) / (-9.8)

= (-6 ± 15,716^0.5) / 9.8

(-6 - 15,716^0.5) / 9.8 would give a negative answer which is not possible in this context.

So the answer is (-6 + 15,716^0.5) / 9.8

Rounded to 2 decimal digits, this is:

12.18 seconds