Question

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.34 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level.

Answer 1

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$

$\hat{ECD} = tan^{-1} 2 = 63.43^o$

This is also the angle ACB, so we can find the length AB:

$tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}$

$2 = \frac{AB}{234}$

$AB = 2*234 = 468 m$

So the height of the building is 468m