Question

A stone is dropped from a balloon ascending with velocity of 9.8 m/s.it reaches ground in 11 sec.How high was the balloon when the stone was dropped and with what velocity did it hit the ground?

Answer 1

(a) u=-5 m/sg=9.8 m/s2S=300 mS=ut+12gt2300=-5t+12×9.8t24.9t2-5t-300=0t=--5±-52-4×4.9×-3002×4.9t=5±25+58809.8t=5±76.849.8=8.35 s ignore the negative value(b) u=5 m/sg=9.8 m/s2S=300 m300=5t+12×9.8t24.9t2+5t-300=0t=-5±52-4×4.9×-3002×4.9t=-5±25+58809.8t=-5±76.849.8=7.33 s ignore the negative value(c) u=0g=9.8 m/s2S=300 m300=0×t+12×9.8t2t2=3004.9t=3004.9=7.8 s

Answer 2

V=9.8m/s
t= 11s
a=g=9.8m/s2
Let the distance be 'x'
Use the equation :
S=Ut +1/2 at^2