To determine when and at what height the two bodies will meet, we need to analyze their respective motions.
Let's consider the body thrown upwards first. We can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
where: h is the height from the ground, u is the initial velocity, g is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time.
For the body thrown upwards: h1 = 40t - (1/2)(9.8)t^2
Now let's consider the body dropped from a height. The equation for its motion is simpler since the initial velocity is 0:
h2 = (1/2)(-9.8)t^2
To find when the two bodies will meet, we equate h1 and h2 and solve for t:
40t - (1/2)(9.8)t^2 = (1/2)(-9.8)t^2
Multiplying both sides by 2 to eliminate fractions:
80t - 9.8t^2 = -9.8t^2
Simplifying:
80t = 0
t = 0
This means the two bodies will meet at t = 0 seconds, which is the initial time when they were both released.
To find the height at which they meet, we substitute t = 0 into either h1 or h2:
h1 = 40(0) - (1/2)(9.8)(0)^2 = 0
Therefore, the two bodies will meet at a height of 0 meters from the ground.
In other words, they will meet at the ground level.