A balloon going upward with a velocity of `12m//s` is at a height of `65m` from the earthh surface at any instant. Exactly at this instant a packet drops from it. How much time will the packet take in reaching the surface of earth?(Take `g=10m//s^(2)`)

A balloon is going upwards with a velocity of 20 m It releases a packet when it is at a height of 160 m from the ground. The time taken by the packet to reach the ground is

A balloon is going vertically upward with a velocity of 15 ms^-1 . When it is at a height of 50 m above the ground a stone is gently dropped from it. The stone will reach the ground in time (take g = 10 ms^-2) :

A balloon is at a height of 40 m and is ascending with a velocity of 10 "m.s"^(-1) . A bag of 5 kg weight is dropped from it. When will the bag reach the surface of the earth? Given g = 10 "m.s"^(-2) .

A balloon is going upwards with velocity 12 m//sec it releases a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground (g=10 m//sec^(2))

A balloon is going upwards with velocity 12 m//sec it releases a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground (g=10 m//sec^(2))

A packet is dropped from a balloon that is moving upward when the balloon is at a height 60 m above ground. If the speed of the balloon at the moment of release of packet is 5 m/s, time taken by the packet to reach ground will be: (Take g = 10 m//s^(2) )