# A pilot flies horizontally at 840 km/h, at height h = 48 m above initially level ground. However, at time t = 0, the pilot begins to fly over ground sloping upward at angle θ = 5.9° ?

##### 1 Answer

#### Answer:

The plane will hit the hill in

#### Explanation:

**FULL QUESTION**

*A pilot flies horizontally at* *at height* *above initially level ground. However, at time* *the pilot begins to fly over ground sloping upward at angle*

*If the pilot does not change the airplane's heading, at what time t does the plane strike the ground?*

The idea here is that you can use the angle the slope makes with the horizontal to determine the *distance the plane covers* before hitting the side of the hill.

Assuming that the *horizontal speed* of the plance does not change, you can then use this distance to find the time it takes the plane to hit the hill.

So, at

If it covers a distance **after** this point in time, then you can use

#tantheta = h/d" "# , where

This means that the plane will travel

#d = h/tantheta = "48 m"/tan(5.9^@) = "48 m"/0.10334 = "464.5 m"#

Now, you need to convert its speed from *kilometers per hour* to *meters per second* before moving on

#840color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * "1000 m"/(1color(red)(cancel(color(black)("km")))) * "1 h"/"3600 s" = "233.3 m/s"#

Since the plane's horizontal speed is presumed constant, you have

#v = d/t implies t = d/v#

#t = (464.5color(red)(cancel(color(black)("m"))))/(233.3color(red)(cancel(color(black)("m")))/"s") = color(green)("1.99 s")#