Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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.

Sagot :

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

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.