To determine when and at what height the two bodies will meet, we can calculate their respective motion equations and find the point of intersection.
Let's consider the first body, which is thrown upwards with an initial velocity of 40 m/s. The acceleration due to gravity (g) will act in the opposite direction, causing the body to decelerate until it reaches its highest point. At the highest point, the body's vertical velocity will be 0 m/s.
Using the equation of motion for displacement in the vertical direction:
s = ut + (1/2)at^2
Where: s = displacement (height from the ground) u = initial velocity a = acceleration (in this case, -g due to deceleration) t = time
At the highest point, the body's velocity is 0 m/s, so we can write:
0 = 40 - 9.8t
Simplifying the equation:
9.8t = 40
t = 40 / 9.8 ≈ 4.08 seconds
Therefore, the first body will take approximately 4.08 seconds to reach its highest point.
Now, let's consider the second body, which is dropped from a height. We can use the equation of motion for displacement in the vertical direction again:
s = ut + (1/2)at^2
For the second body, the initial velocity (u) is 0 m/s because it is dropped. The acceleration (a) is the same as before, -g.
s = (1/2)(-9.8)t^2
Since we know the initial velocity of the second body is 20 m/s, we can set the equation:
20 = (1/2)(-9.8)t^2
Simplifying the equation:
-9.8t^2 = 20 * 2
-9.8t^2 = 40
Dividing both sides by -9.8:
t^2 = 40 / -9.8
t^2 ≈ -4.08
The time (t) squared is negative, which indicates that there is no real solution. This means that the second body does not meet the first body in the given scenario.
In conclusion, the two bodies will not meet each other in the air.