A man bailed out of a balloon. After sometime the parachute opened up and he could land on the earth's surface at a retardation of 2.4 m. s-2, and took 4 times the time that elapsed before the parachute opened. If the ballon  was at a height of 398.4 m, how long was the airborne? At what speed did he touch the ground?

Dear student

The figure shows the situation 


At A when man bailed of the balloon he fall of under gravity so 
from A to B: let at be the velocity of man is v then 
v=u+gtv=0+gtv=9.8t                                            .......(1)and398.4-y=0+12gt2398.4-y=9.82t2398.4-y=4.9t2                      ..........(2)
From B to C:
let the velocity of man when he touches the ground is v' then
v'=v+a·4tv'=9.8t-2.4×4tv'=9.8t-9.6tv'=0.2t                            ......(3)and y=v·4t+12a4t2y=9.8t·4t-12×2.4×16t2y=39.2t2-19.2t2y=20t2but from (2)398.4-4.9t2=20t2t2=398.424.9t2=16t=4 sec
Thus v'=0.2×4=0.8 m/s2 andy=20×16=320m
 

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