Question

A rocket carrying fireworks is launched from a hill 80 feet above a lake. The rocket will fall into the lake after exploding at its maximum height. The rockets height above the surface of the lake is given by the function d(t) - 16t2 – 14t + 382. 52. What is the height of the rock after 3 seconds?

Answer 1

The height of the rocket after 3 seconds is 196 feet above the surface of the lake, which is determined by substituting t = 3 into the given height function.

Step-by-step explanation:

To determine the height of the rocket after 3 seconds, we need to plug t = 3 into the height function d(t) = -16t2 - 14t + 382. Performing the calculation:

• First, calculate the term with t squared: -16(3)2 = -16(9) = -144.
• Then, calculate the term with t: -14(3) = -42.
• Add these to the constant term: -144 - 42 + 382.
• Combine these to find d(3), which is the height of the rocket after 3 seconds.

The calculation yields:

-144 - 42 + 382 = 196

Therefore, the height of the rocket after 3 seconds is 196 feet above the surface of the lake.