Solve this please and explain how you do it thank you

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

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

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

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

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

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

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

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

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

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

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

3 years ago

1:2000000 find the land distance 14cm b) 31 cm

bogdanovich [222]

If the scale is

R.F.= 1/2000000

=1/2x 10000x100

1cm=20000m

The scale of a map is given as= 1/2000000

Two town are 4 cm apart on the map.

the actual distance between two towns

=4x 20000

=80000 m

=80000/1000 Km

=80Km

I hope this help you

6 0

3 years ago

Nurse Jackson examines immunization records for this past winter at the school where she works. Got flu Didn't get flu Total Vac

Harman [31]

Answer:

140

Step-by-step explanation:

ap3x

6 0

2 years ago

Which pair of measurements is not equivalent?

mamaluj [8]

D is not equivalent
hope this helps :)

5 0

2 years ago

Read 2 more answers

One circular table has a diameter of 8 ft, and another circular table has a diameter of 12 ft. How much greater is the area of t

mezya [45]

Answer:

20π sq ft ≈ 62.83 sq ft

Step-by-step explanation:

Area of table with diameter 8 ft = π*8^2/4 = 16π

Area of table with diameter 12 ft = π*12^2/4 = 36π

Difference between the area of the tables ;

= 36π - 16π

= 20π sq ft ≈ 62.83 sq ft ( π = 3.14)

8 0

3 years ago

Read 2 more answers