A particle is moving up with a ballon with constant acceleration g/8 which starts from rest from ground and at heightH particle is dropped from the balloon after this event find the time particle will be in air

To solve this problem, we need to analyze the motion of the particle and the balloon separately. The balloon is moving upward with a constant acceleration, and once the particle is released, it will be influenced by gravity. Let's break this down step by step.

Understanding the Balloon's Motion

The balloon starts from rest and accelerates upwards with an acceleration of \( g/8 \). Here, \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \). Therefore, the acceleration of the balloon is:

a = g/8 = 9.81/8 \approx 1.22625 \, \text{m/s}^2

Calculating the Height of the Balloon

Since the balloon starts from rest, we can use the kinematic equation for distance traveled under constant acceleration:

H = ut + (1/2)at²

Here, \( u = 0 \) (initial velocity), \( a = g/8 \), and \( t \) is the time taken to reach height \( H \). Thus, the equation simplifies to:

H = (1/2)(g/8)t²

Rearranging gives us:

H = (g/16)t²

Analyzing the Particle's Motion After Release

Once the particle is dropped from the balloon at height \( H \), it will have an initial velocity equal to the velocity of the balloon at that moment. We can find this velocity using the equation:

v = u + at

Since the initial velocity \( u = 0 \) and the acceleration is \( g/8 \), the velocity of the balloon at height \( H \) is:

v = (g/8)t

Time of Flight for the Particle

After being released, the particle will experience free fall, with an initial upward velocity \( v \) and an acceleration of \( -g \) (downward). The time \( t_f \) it takes for the particle to hit the ground can be found using the kinematic equation:

y = vt_f + (1/2)(-g)t_f²

Setting \( y = -H \) (since it falls to the ground), we have:

-H = (g/8)t_f t + (1/2)(-g)t_f²

Substituting \( H \) from earlier:

- (g/16)t² = (g/8)t_f t - (1/2)gt_f²

Multiplying through by -1 and rearranging gives:

(1/2)gt_f² - (g/8)t_f t - (g/16)t² = 0

Now, we can simplify this equation by dividing everything by \( g \):

(1/2)t_f² - (1/8)t_f t - (1/16)t² = 0

Solving the Quadratic Equation

This is a standard quadratic equation in the form \( at^2 + bt + c = 0 \), where:

• a = 1/2
• b = -1/8 t
• c = -1/16 t²

Using the quadratic formula \( t_f = \frac{-b \pm \sqrt{b² - 4ac}}{2a} \), we can find \( t_f \):

t_f = \frac{(1/8)t \pm \sqrt{(-1/8 t)² - 4(1/2)(-1/16 t²)}}{2(1/2)}

Calculating the discriminant and simplifying will yield the time of flight \( t_f \). The positive root will give us the time the particle is in the air after being released from the balloon.

In summary, the total time the particle is in the air consists of the time it takes to reach height \( H \) plus the time it takes to fall back to the ground after being released. By carefully applying kinematic equations, we can solve for both components of the motion effectively. If you have any further questions or need clarification on any steps, feel free to ask!