1)a ballon is asending at the rate of 14m/s at height of 98m above the ground, when a packet is dropped from the ballon.AFter how much time and with what velocity it will reach the ground

2)if at the moment t=0,the seperation between the cars is 100m,the seperation between them at the moment t=2 will be.

1) Initially the ball is going upward
   u=-14 m/s, s=98 m, a=g=10 m/s2s=ut+12at298=-14t+12×10t298=-14t+5t25t2-14t-98=0t=14±196-4×5×(-98)2×5t=14±46.4310

taking positive sign

t=14+46.4310=6.043 s  (t never be negative)

Now the velocity at which the packet reached to the ground,
           v=u+at   =0+10×6.043   =60.43 m/s

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