a body is projected with velocity v1 from a point a at the same time another body is projected vertically upwards with velocity v2 the point b lies vertically below the height the point for both the bodies to collide 2/v should be

Why did the body and the point decide to collide? Because they couldn't resist each other's attraction! So, to calculate the time it takes for them to collide, we need to find the height at which they meet.

Let's assume that the height is h. The body projected from point A will travel a horizontal distance of 2v (since it has a velocity of v1) and the body projected vertically upwards will reach a height of h (since it has a velocity of v2).

Now, let's think of these two bodies as simultaneous acrobats performing a daring stunt. The body from point A will take time t1 to travel the horizontal distance of 2v, and the body projected upwards will take time t2 to reach a height of h.

Since they are performing together, the time taken by both bodies should be the same. Therefore, we can say that t1 = t2.

We know that time can be represented as distance divided by velocity. So, t1 = (2v) / v1 and t2 = h / v2.

Setting these two equal to each other, we get (2v) / v1 = h / v2.

Now, let's flip this equation around to make it more pleasing: 2/v = (h * v1) / (v2).

So, 2/v = (h * v1) / (v2).

And thereby, my friend, we have found the magical equation that will help us determine the value of 2/v in terms of h, v1, and v2.

To find the time it takes for the bodies to collide, we need to consider the vertical motion of both bodies.

Let's assume:

1) The initial velocity of the body projected from point A is v1.
2) The initial velocity of the body projected vertically from point B is v2.
3) The time taken for both bodies to collide is t.
4) The height between points A and B is h.

For the vertically projected body from point A:
Using the equation of motion for vertical motion:
h = v1t - (1/2)gt^2 -------(1)

For the vertically projected body from point B:
Using the equation of motion for vertical motion:
h = v2t + (1/2)gt^2 -------(2)

Adding equations (1) and (2) to eliminate h:
2h = (v1 + v2)t -------(3)

Now, to find the time it takes for both bodies to collide, we need to rearrange equation (3):

t = 2h / (v1 + v2)

dividing both the numerator and denominator by 2v:
t = h / (v1/2 + v2/2)
t = h / ((v1 + v2)/2)

Therefore, 2/v = (v1 + v2)/2

Hence, 2/v should be equal to (v1 + v2)/2.

To find the time it takes for the two bodies to collide, we need to analyze their motion separately and then set up an equation equating their positions.

Let's assume that the body projected from point A travels horizontally and the body projected vertically travels only in the vertical direction (upwards) from point B.

For the body projected horizontally from A:
Since there are no forces acting in the horizontal direction, its horizontal velocity, v1, remains constant throughout its motion. Therefore, the horizontal distance covered by the body can be given as d = v1 * t, where d is the horizontal distance and t is the time taken for the collision.

For the body projected vertically from B:
We can use the equation of motion to determine the height reached by the projectile. The equation is given by:
h = v2 * t - (1/2) * g * t^2, where h is the height, v2 is the initial vertical velocity, g is acceleration due to gravity (9.8 m/s^2), and t is the time taken for the collision.

Since the collision occurs at the same height, the height reached by the vertically projected body should be equal to the vertical distance to point B. Therefore, h = h, and v2 * t - (1/2) * g * t^2 = h.

To simplify the equation, we can substitute the value of h with the vertical distance to point B, which we'll call "x". Thus, v2 * t - (1/2) * g * t^2 = x.

Now, we can solve for t by rearranging the equation:
(1/2) * g * t^2 - v2 * t + x = 0.

This quadratic equation can be solved using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = (1/2) * g, b = -v2, and c = x.

Simplifying the equation further:
t = (-(-v2) ± sqrt((-v2)^2 - 4 * (1/2) * g * x)) / (2 * (1/2) * g)
t = (v2 ± sqrt(v2^2 - 2 * g * x)) / g

Now, we have the time t in terms of v2 and x.

But the question asks for the value of 2/v. Let's use the equation d = v1 * t to find the value of v.

v1 = d / t
v1 = v * t (since d = v1 * t and the horizontal velocity remains constant)
v = v1 / t

Therefore, 2/v = 2 / (v1 / t) = 2t / v1.

Now, let's substitute the value of t we found earlier:
2/v = 2 * [(v2 ± sqrt(v2^2 - 2 * g * x)) / g] / v1

So, 2/v is equal to 2 times the expression [(v2 ± sqrt(v2^2 - 2 * g * x)) / g] divided by v1.

Note: The expression in the square root, v2^2 - 2 * g * x, should be greater than or equal to zero for the two bodies to collide.