Answer to Question #180468 in Atomic and Nuclear Physics for ary

Question

1. An observer stands 600 meters from the launch pad of a rocket, watches it       ascend vertically at 100 m/s.

a.  Find the rate of change of the distance between the rocket and the observer. (Hint: Use Pythagorean Theorem)

 b.    Find the distance between the observer and the rocket when the rocket is 400 meters high.

  

 c.   What is the rate of change of the distance between the rocket and the observer when the rocket is 400 meters high?

Answer 1

To be given in question

Observer and rocket between distance  =600meter

Vertically velocity of rocket =100m/sec

To be asked in question

Rate of change distance between observer and rocket $\frac {dh}{dt}=?$

Solution (a)

$x^2+y^2=h^2$

Take differentiate both side

$2y \frac{dy}{dt}=2h\frac{dh}{dt}$

$y \frac{dy}{dt}=h\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{y}{h}\frac{dy}{dt}$

$\frac {dh}{dt}$ $=\frac {500}{\sqrt {500^2+600^2}}\times100$

$\frac {dh}{dt}=64.02meter/sec$

Solution (b)

When Y=400meter/sec

Rocket and observer between distance pithagorse theorem

Use

$h^2=x^2+y^2$

$h^2=600^2+400^2$

$h=√5200$

h=72.11meter

Solution (c)

Pauthagorse theorem

$x^2+y^2=h^2$

Take differenciate 

$y \frac{dy}{dt}=h\frac{dh}{dt}$

$2y \frac{dy}{dt}=2h\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{y}{h}\frac{dy}{dt}$

$\frac{dh}{dt}=\frac{400}{\sqrt{600^2+400^2}}\times100$

$\frac{dh}{dt}=55.44meter/sec$