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Question: A balloon starts rising from the ground with an acceleration of 1.25\(m{s^{ - 2}}\).After 8s, a ston...

A balloon starts rising from the ground with an acceleration of 1.25ms−2m{s^{ - 2}}.After 8s, a stone is released from the balloon. The stone will (taking10ms−210m{s^{ - 2}}).
A. Begin to move down after being released.
B. Reach the ground in 4s
C. cover a distance of 40m in reaching the ground
D. Will have a displacement of 50m.

Explanation

Solution

The equations which established the relation between initial velocity, final velocity, time, acceleration, and displacement are known as the kinematic equation of motion. On releasing the stone from the balloon it initially has upward velocity.
Formula used:
Displacement of the balloon, s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
The velocity of the balloon, v=u+atv = u + at
Where,
s = displacement (m)
u = initial velocity (m/s)
v =the final velocity (m/s)
a =the acceleration of the body (ms−2m{s^{ - 2}} )
t = the time is taken (s)

Complete step-by-step answer:
We are Given with, a balloon starts rising from ground therefore initial velocity u=0 m/s
As it moves upwards its acceleration will be 1.25ms−2m{s^{ - 2}} and it reaches a certain height then a stone falls from the balloon after 8s.
i.e. u=0 m/s, a=1.25ms−2m{s^{ - 2}} , t=8s
Now we will calculate the distance of the stone above the ground about which it begins to fall from the balloon.
Here, Let s= h, u=0m/s, and a=1.25ms−2m{s^{ - 2}}, t=8s
Then, substituting the values in the formula s=ut+12at2s = ut + \dfrac{1}{2}a{t^2} we get,
h=0+12(1.25)82h = 0 + \dfrac{1}{2}\left( {1.25} \right){8^2}
Therefore, h=40m.
Stone covers a distance of 40m before reaching the ground.
Next, we will find out the velocity of stone at height 40m,
The velocity of the balloon at this height can be obtained from the formula v=u+atv = u + at
Here, u=0, a=1.25ms−2m{s^{ - 2}}, t=8s
Then we get,
v=0+(1.25)8=10ms−1v = 0 + (1.25)8 = 10m{s^{ - 1}}.
This velocity becomes the initial velocity (u’) of the stone as the stone falls from the balloon from height h.
Therefore we have now, the initial velocity at height h, u’= 10ms−110m{s^{ - 1}}
Now let us calculate the total time taken by the stone to reach the ground.
Total motion of the stone, is given by an equation,
h=12gt2−u′th = \dfrac{1}{2}g{t^2} - u't
Here, h=40m, u’= 10ms−110m{s^{ - 1}}and t is the time taken by the stone to reach the ground.
Substituting values in kinematic equation, we get
−40=10t−12×10t2- 40 = 10t - \dfrac{1}{2} \times 10{t^2}
After solving,
−40=10t−5t2- 40 = 10t - 5{t^2}
5t2−10t−40=05{t^2} - 10t - 40 = 0
Divide the above equation by 5, we get
t2−2t−8=0{t^2} - 2t - 8 = 0
After factorization,
t2−4t+2t−8=0{t^2} - 4t + 2t - 8 = 0
t(t−4)+2(t−4)=0t\left( {t - 4} \right) + 2\left( {t - 4} \right) = 0
Therefore t-4=0 or t+2=0
Then the value of t comes to be, t=4 or -2
Ignoring the negative value of the time we get, t=4s.
Therefore, the correct option is (B).
Additional information:
Selection of kinematic equation: Choose the formula which includes both the unknown variables looking for and three of the kinematic variables we already know. Such that, we can solve the unknown which will be only unknown in the formula.

Note:
The acceleration due to the gravity of a body is always directed downwards toward the center of the earth, whether a body is projected upwards or downwards.
When a body is falling towards the earth, its velocity increases, g is positive
When a body is projected upwards, its velocity decreases, g is negative