Question

The height, in feet, of a toy rocket that is launched from 4 feet above the ground can be modeled using the equation provided where t is the time after the rocket is launched. To the nearest foot, what is the height of the rocket 3 seconds after it is launched? ​

Answer 1

Unfortunately, there is no equation provided in the question. However, I can provide a general formula for the height of a rocket launched from a certain height h0 with an initial velocity v0, under the influence of gravity:

h(t) = -1/2gt^2 + v0t + h0

where g is the acceleration due to gravity, which is approximately 32.2 feet per second squared.

Using this formula, we can calculate the height of the rocket 3 seconds after it is launched:

h(3) = -1/2(32.2)(3)^2 + 0 + 4

= -1/2(32.2)(9) + 4

= -145.35 + 4

= -141.35

To the nearest foot, the height of the rocket 3 seconds after it is launched is -141 feet. Note that the negative sign indicates that the rocket has fallen below its initial height of 4 feet due to the influence of gravity.

Answer 2

). The rocket was launched from 73 feet above ground.

b). The rocket took [Vy / g] = 30 / 3 = 10s to reach its maximum height of [Vy^2 / (2 * g)] = 30^2 / 6 = 150 feet.

It then fell -(73 + 150) = -223 feet to the ground in a time of sqrt(2 * -223 / -3) = 12.192894s.

Total time of flight = (10 + 12.192894) = 22.192894s.

The rocket landed with a vertical velocity of -sqrt(2 * -223 * -3) = -36.578682ft/s.

Explanation: