a balloon starts ascending upwards with a constant velocity of 10 m/s when it has reached a height of 40 m above the ground the packet of 2 kg is dropped from it .The time after which it strikes the ground is( g=10m/s2)
Answers
You can use the Equations we use to solve linear motion. Here, when the 2Kg packet drops from the balloon, Since the balloon has a velocity of 10m/s upwards, 2 Kg packet also has an initial velocity of 10 m/s upwards. And the g is given in the question. g will always be towards the ground. So we have below information regarding the Balloon when it dropped from the balloon,
U = -10 m/s, a = g = 10 m/s2, S = 40 m, t = ?
In order to find t, we can use linear motion equation S = ut + 1/2at^2
we wanted apply this equation towards ground,
40 = - 10 x t + (1/2) x 10 x t^2
4 = -t + (1/2) x t^2
8 = -2t + t^2
t^2 - 2t - 8 = 0 by solving this quadratic equation you will get below result,
(t-4)(t+2) = 0
t = 4 or t = -2
Since time can not be negative t = 4 Seconds.