Question

A hot-air balloon of mass 500 kg is descending at a speed of 5 ms^-1 with an acceleration of 0.2 ms^-2.

(i) What mass of ballast must be thrown out to reduce the acceleration to zero?

(ii) If twice this amount of ballast is thrown out, how much further will the balloon descend before it starts to climb?

Answers: (i) 10 kg (ii) 60 m

Answer 1

o ifm is the mass y balloon and it is descending down with acceleration f. then From 2nd law Newtas the equation 1 to tif 7 g Fince Ting mg - F = M f - is where for is the thoyant force. - O رال - Hw if mass im thronon nt then acceleration belome zero. now equalia gi Forced. (M-mig - FB FB = (m-myg M-m) fco put this in equis mg-(m-myg=mt (M-mog mg-ng ting=mf m=mf g We have given M = 50kg f= 0.2 m/s² and put g = 10 m/s²

m- 510 x 0:2 16 tokg m=lokg From ean (ii) Fo = (m-mig FR = 1500-10) X10 = 4900 N. med at (M-2mJg whene am mass is thrmon out After travelling s distance it start to climb then equation og fance F-(m.2m) g = (M-2m) og 4900-(510-2810) X10 = (500-2X10 )f' 49o 4800 = 48of! To = 480 ft f'=100 10 m/s² 480 48 so distance by equalia, of motin U²=u²-2f's = n²-24's S= 42 5X5X46 2x 10 Gom 2f'