Question

WILL GIVE BRAINLIEST A rocket is launched vertically from the ground with an initial velocity of 64. Write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched. Graphon the coordinate plane.

Answer 1

Answer:

$h=64t-4.9t^{2}$

Please refer to the attached graph.

Step-by-step explanation:

Given that

Initial velocity of rocket = 64 and is launched vertically.

To find:

Quadratic equation in time to represent the height of rocket in feet.

Solution:

Unit of initial velocity is not given in the question statement, let the velocity be in feet/second only.

Initial velocity, u  = 64 feet/s

The acceleration will be = -g because it is going opposite to gravitational force so it will be negative acceleration motion (speed will be decreasing) so -g will be the acceleration.

Formula for distance traveled is given as:

$s=ut+\dfrac{1}{2}at^2$

Here Let us represent s by 'h'

a = -g = 9.8 m/$s^2$

Let us put the known values in the formula:

$h=64t+\dfrac{1}{2}(-9.8)t^2\\\Rightarrow h =64t-4.9t^2$

It is a quadratic equation, the equation represents the graph of a parabola.

Please refer to the attached graph.

Value of height is 0 at 0 second and ~13 seconds