A balloon rises up with constant net acceleration of 10 ms^2, starting from rest. After 2s a particle drops from the balloon. What is the maximum height attained by the particle?



Using the formula of motion for constant acceleration, we can calculate the height of the balloon after 2 seconds.


s = ut + (1/2)at^2

where,
s = distance traveled
u = initial velocity (0 m/s for the balloon)
a = acceleration (10 m/s^2 for the balloon)
t = time (2 seconds for the balloon)


s = 0 + (1/2) x 10 x 2^2
s = 20 meters

Therefore, the height of the balloon after 2 seconds is 20 meters.


As the particle is dropped from the balloon, it will have an initial velocity of 0 m/s and will accelerate downwards with an acceleration of 9.8 m/s^2 (acceleration due to gravity).


s = ut + (1/2)at^2

where,
s = distance traveled (20 meters from the balloon)
u = initial velocity (0 m/s for the particle)
a = acceleration (-9.8 m/s^2 for the particle)
t = time taken to reach the ground


20 = 0 x t + (1/2) x (-9.8) x t^2
t^2 = 4.08
t = 2.02 seconds (approx)

Therefore, the time taken by the particle to reach the ground is approximately 2.02 seconds.


To calculate the maximum height attained by the particle, we need to find the distance traveled by the particle before hitting the ground.


s = ut + (1/2)at^2

where,
s = distance traveled
u = initial velocity (0 m/s for the particle)
a = acceleration (-9.8 m/s^2 for the particle)
t = time taken to reach the ground (2.02 seconds)


s = 0 x 2.02 + (1/2) x (-9.8) x (2.02)^2
s = 20.2 meters (approx)

Therefore, the maximum height attained by the particle is approximately 20.2 meters.


The maximum height attained by the particle dropped from the balloon after 2 seconds is approximately 20.2 meters.

The maximum height is 40m as let in 2 sec height attain (H) that is equal to ut+1/2at2 where u=0 so H=1/2*10*2*2on solving H=20m now final velocity of balloon after 2 sec will be V=u+at where u=0 so V=10*2=20m/s. now this velocity will be transfer to particle so u of particle=20m/s and a=-10m/s2 and v=0 for max height so by 3rd eqn of motion v2=u2+2as. on putting values 0=400-2*10*s and s=20m. (on solving) where s=height attain after stone is relesed from balloon so total max height from ground will be s+H that is 20+20=40m that's mine answer...