geometry sum

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!

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

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 

Very nice diagram, reinout.......!!!!!!

Yes, it is a very nice diagram and I think that you should give reinout some more points Chris. !!!     LOL

Geogebra will not open for me so at the moment I am diagram-less