Question

a balloon start rising upwards with constant acceleration A and after T second a packet is dropped from it which reaches the ground in t" . the value of t" is _​

Answer 1

Step-by-step explanation:

Let us assume that the balloon starts rising from ground level with constant acceleration a.

We calculate the height h attained in time t_1 using the kinematic expression

h = ut + ½ a t^2

h = ½ a t_1^2 ……..(1)

Velocity at this height is calculated with the kinematic expression

v = u + at

v = at_1 ……….(2)

When the packet is dropped from this height, it reaches ground in time t. Neglecting air friction we get

h = a t_1 t - ½ 9.8 t^2

Inserting value of h from expression (1) we get

½ a t_1^2 = a t_1 t - ½ 9.8 t^2

rewriting it we get a quadratic expression in t

9.8 t^2 - (2a t_1) t + (a t_1^2) = 0

Solving for t and rejecting negative root we obtain the desired value

Answer 2

Analysis of situation: t = 0 is the time when the balloon started rising up. At t = 0, when =

the packet is dropped, the balloon is moving up with velocity v = 0 + ato = ato. Hence, initial

VO velocity of the packet will be vp = ato

(upward). As the balloon has started rising upwards with constant acceleration a, so

after to seconds, its height from the groundis

is yo = atz.

For packet: s = ut-1/2gt^2

-1/2atot-1/2gt^2=gt^2 -2atot -ato^2=0

Solving the quadratic equation, we get t =

9.8 t^2 - (2a t_1) t + (a t_1^2) = 0