Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A radar station detects a missile approaching from the east. At first contact, the range to the missile is 12,000 ft at 40.0° above the horizon. The missile is tracked for another 123° in the east-west plane, the range at final contact being 25,800 ft; see Fig. Find the displacement of the missile during the period of radar contact.

A radar station detects a missile approaching from the east. At first contact, the range to the missile is 12,000 ft at 40.0° above the horizon. The missile is tracked for another 123° in the east-west plane, the range at final contact being 25,800 ft; see Fig. Find the displacement of the missile during the period of radar contact.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
It is assumed that the position vector of the missile when it was first located is given by vector\overrightarrow{a} making an angle \theta _{1} with the horizontal axis pointing east, in the direction of the unit vector\widehat{i} , measured counterclockwise.
When the missile was later located, its position vector is given by vector \overrightarrow{b}, making an angle\theta _{2} with the horizontal axis pointing west, in the direction opposite to the unit vector\widehat{i} , measured clockwise.
Also, the magnitude of vector\overrightarrow{a} is given by a whereas the magnitude of vector\overrightarrow{b} is given by b .
Given:
236-2128_eq.PNG
The figure below shows the position vectors of the missile relative to radar.

236-469_now.PNG

Position vector\overrightarrow{a} for the missile is given as:
\overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j}
Where ax and ay represents the horizontal and vertical components of the vector.
Also,a_{x} = a\ cos\theta _{1} and a_{y} = a\ sin\ \theta (refer figure above).
Therefore position vector\overrightarrow{a} can be written as:
\overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j}
\overrightarrow{a} = a\ cos\theta _{1}\widehat{i}+ a\ sin \theta_{1}\widehat{j}
Substitute the given values of a and \theta _{1} in the vector above to have
236-2452_eq2.PNG
Therefore the position vector of the missile when it was first located by the radar is
236-1960_eq3.PNG
Position vector\overrightarrow{b} for the missile is given as:
236-1946_eq4.PNG
Where bx and by represents the horizontal and vertical components of the vector.

Also,b_{x} = b\ cos\ \theta_{2} and b_{y} = b\ sin\ \theta _{2} (refer figure above).

The value of\theta _{2} can be calculated as
\theta _{2} = 180°-123°-40°
= 17°
Therefore position vector \overrightarrow{b}can be written as:
236-1126_eq5.PNG
The negative sign in horizontal component above highlights the fact that the vector component points in the direction opposite to unit vector \widehat{i}.

Substitute the given values of b\ and\ \theta _{2} in the vector above to have
236-1454_eq6.PNG
Therefore the position vector of the missile when it was located later by the radar is
.236-1360_eq7.PNG
The displacement vector (say \overrightarrow{r}) of the missile is given as:

236-1755_eq8.PNG
If it is assumed that the displacement vector of the missile is given as:
236-1519_eq9.PNG
Then on comparing the above vector with vector in equation (1), we have
rx = -15, 480.1 ft
ry = -170.3 ft
The magnitude of displacement of missile is given as:
236-2104_eq10.PNG
Substitute the value of rx and ry from above

236-1013_eq11.PNG
Rounding off to two significant figures, we have
|\overrightarrow{r}| = 15,000\ ft
Therefore the magnitude of displacement vector of the missile is 15,000 .

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free