You are in a timed hot-air balloon race, cruising at a constant height of 1km ,

Question

You are in a timed hot-air balloon race, cruising at a constant height of 1km, and with a constant velocity of 15km/h. The race will end in 2 minutes, at which point the balloon must still be airborne or you will be disqualified. Suddenly the temperature in the balloon drops and so does the buoyancy. It now provides only 90% the lift it did before (that is , Fnow=0.9Fbefore). Unfortunately, your air heater is broken, but you do have a secret rocket propulsion that can fire a thrust, as shown. If the rocket fired directly own (providing an upward thrust) it would exactly compensate for the lost buoyancy. However, in order to win the race, you need to travel an additional 2km, and a quick calculation shows that you won't make it at your current velocity. At what angle should you point the rocket jets to ensure that you win the race, without hitting the ground first? Ignore the effects of air resistance.

Explanation / Answer

I first got that Fnow=0.9Fbefore=0.9mg and Frocket=0.1Fbefore=0.1mg. I then drew the force body diagram and in the x direction is Frocketcos=max and in the y direction is Frocketsin+Fnowmg=may and after plugging in the values and cancelling the mass out, I got for the x direction 0.98cos=ax and the the y direction 0.98sin0.98=ay

You just need the x-component of the acceleration because the sine of the angle made between the weight and the rocket thrust is a_x/mg.

As a_x = 0.208 m/s from the displacement equation the required angle is 12.24 degrees.

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