# A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

9 years ago
Sol. Initially the ball is going upward u = –7 m/s, s = 60 m, a = g = 10 m/s2 s = ut+ ½ at2 ⇒ 60 = -7t + ½ 10t2 ⇒ 5t2 – 7t – 60 = 0 t = (7+ √(49-4.5 (-60) ))/(2 x 5) = ( 7 ± 35.34 )/10 taking positive sign t = (7+ 35.34)/10 = 4.2 sec (∴ t ≠ -ve) Therefore, the ball will take 4.2 sec to reach the ground.
Arvind
19 Points
6 years ago
Its ans is 3.49s SolnBallon is at the distance of 60m from the ground and a ball is dropped from it so intial velocity will be zero and a=g=9.8ms^2 and they are asking time taken by the ball to reach the ground So by applying s=ut+1/2gt^260=0+1/2×9.8×t^260=4.9t^2T^2=60÷4.9T^2=12.24T=3.49s
Guy
11 Points
6 years ago
If we look from the frame of balloon the ball is always going down with an initial velocity 7m/s ,with acceleration 9.8m/s^2. Therefore, the final velocity of ball is-
v^2=u^2+2as = 49+2*9.8*60. v^2=1225. v=35m/s.
now,time is, v=u+at. =>t=(v-u)/a.t= (35-7)/9.8. =>t= 2.857 seconds.
Devansh mamoriya
13 Points
6 years ago
As ball is released from ballon than firstly it move upward hen it`s velocity become zero it fall on the ground when we take observer from ground and then using formula s=ut+1/2at^2 with appropriate sign convention we can find the answer