A balloon goes upward to 98 m at a speed of 9.8m/s. Then one particle is released from the balloon. What is the time taken by the particle to...

In this case, since the balloon is moving upwards at 9.8 m/s, the particle will also be ejected at the same speed. This particle will travel upwards, reach a maximum height and then start falling downwards.

For the upwards motion, the final velocity will be zero.

Thus, v = u + at 

or, 0 = 9.8 + (-9.8) t

solving the equation, we get t = 1 s.

That is, the particle will travel upwards for 1 second and then start falling down.

During this 1 s, the particle will travel some distance. Using the equation of motion:

v^2 = u^2 + 2as

0^2 = 9.8^2 + 2 (-9.8) s

we get, s = 4.9 m.

Thus the particle will reach a maximum height of 98 m + 4.9 m = 102.9 m before start falling down.

For the downward motion, the initial velocity, u = 0 m/s

Using the equation of motion: s = ut + 1/2 at^2, we get

102.9 = 0 x t + 1/2 (9.8) t^2

solving the equation, we get, t = 4.58 s

Thus, the particle will need a total of 5.58 s (1 s for upward motion and 4.58 s for downward motion).

Hope this helps.