Question

A rock is dropped off a cliff that is 18 meters high into a lake below. (Assume there is no air resistance.) The formula for the height of an object dropped is h(t)= 1/2at^2 +s where the gravitational constant, a, is -9.8 meters per second squared, s, is the initial height, and h(t) is the height in meters modeled as a function of time, t. Which of the following are true select all that apply.
The equation for the situation is h(t)= 4.9t^2 + 18.
The equation for the situation is h(t)= -4.9t^2 + 18.
The rock hits the surface of the water after 1.92 seconds.
The rock hits the surface of the water after 2.02 seconds.

Answer 1

Answer:

2nd and 3rd choices

Step-by-step explanation:

use h(t) = h(0) + v(0)t + at^2/2

Now h(0) = 18, v(0) = 0, a = -9.8

so h(t) = 18 - 4.9 t^2

Solve h(t) = 18 - 4.9 t^2 = 0

Get t = 1.92 sec