Asmall rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 right angle m/s. a cross wind blowing to the north produces an acceleration of the rocket of 2.5 m divided by s squared. assume the x-axis points east, the y-axis points north, the positive z-axis is vertical (opposite g), and the ground is horizontal. answer parts a through d.
find:
a/ velocity and position vectors. b/ determine the time of the flight and range of the rocketc/ determine the maximum hight of the rocket.

Let denote the rocket's position, velocity, and acceleration vectors at time .

We're given its initial position

and velocity

Immediately after launch, the rocket is subject to gravity, so its acceleration is

where .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

and

b. The rocket stays in the air for as long as it takes until , where is the -component of the position vector.

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

c. The rocket reaches its maximum height when its vertical velocity (the -component) is 0, at which point we have

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: