The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation is situated 300 meters away from the take off point. What is the upward speed, assumed constant, of the balloon? Give the answer in meters per second and round to two decimal places.

manas Oct 27, 2015

#3**+10 **

I made a sketch to illustrate your problem.

I divided the question into 4 sub-questions which should lead to your answer.

Try to solve these step by step and see if this solves your problem.

1. How can we use the equation \(tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \) to find height a and b ? (have a look at https://www.mathsisfun.com/sine-cosine-tangent.html if you don't know how this works)

2. How can we use height a and b to calculate the increase in height of the balloon?

3. How much time has passed for the balloon to cover this distance?

4. If we know how much time has passed and how much distance the balloon has covered, how do we calculate the speed?

Good luck!

reinout-g Oct 27, 2015

#1**+10 **

At 10:00 the balloon's height, ( h ) =

tan 25 = h / 300 → h = 300tan 25 = 139.89 m

At 10:02, the height is ;

h = 300tan 60 = about 519.62 m

So in 120 seconds, the balloon rises at

[ 519.62 - 139.89 ] / 120 = about 3.16 m/s

CPhill Oct 27, 2015

#2**+10 **

Use the tangent to write

tan(25o) = h1 / 300

and

tan(60o) = (h1 + h2) / 300

Solve for h1 and h2

h1 = 300 tan(tan(25o))

and

h1 + h2 = 300 tan(60o)

Use the last two equations to find h2

h2 = 300 [ tan(60o) - tan(25o) ]

If it takes the balloon 2 minutes (10:00 to 10:02) to climb h2, the the upward speed S is given by

S = h2 / 2 minutes

= 300 [ tan(60o) - tan(25o) ] / (2 * 60) = 3.16 m/sec

manas Oct 27, 2015

#3**+10 **

Best Answer

I made a sketch to illustrate your problem.

I divided the question into 4 sub-questions which should lead to your answer.

Try to solve these step by step and see if this solves your problem.

1. How can we use the equation \(tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \) to find height a and b ? (have a look at https://www.mathsisfun.com/sine-cosine-tangent.html if you don't know how this works)

2. How can we use height a and b to calculate the increase in height of the balloon?

3. How much time has passed for the balloon to cover this distance?

4. If we know how much time has passed and how much distance the balloon has covered, how do we calculate the speed?

Good luck!

reinout-g Oct 27, 2015