A canyon is 900 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of the rock as a function of time t in seconds. (Use -9.8 m/s2 as the acceleration due to gravity.) How long will it take the rock to hit the canyon floor? (Give your answer correct to 1 decimal place.)

Question
Functions
asked 2021-01-31
A canyon is 900 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of the rock as a function of time t in seconds. (Use -9.8 m/s2 as the acceleration due to gravity.) How long will it take the rock to hit the canyon floor? (Give your answer correct to 1 decimal place.)

Answers (1)

2021-02-01
The height as a function of time h(t) is equal to the initial velocity of the rock v(0) times time (t) plus \frac{1}{2} the acceleration due to gravity (g) times time square (t^{2}). That is, \(h(t) = v(0)*t + 0.5*g*t^{2}\)
Since the rock is dropped and not thrown, there is no initial velocity. Thus, v(0) = 0 and \(h(t) = 0.5*g*t^{2}\)
Assuming the height at the top of the canyon is 0, when the rock hits the canyon floor, h(t) = -900. \(-900 = 0.5(-9.8)t^{2}\)
Solving for t, \((2*900)/9.8 = t^{2}\)
\(t = 13.6 seconds\)
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