To find the location and time when the two balls meet, we need to determine the time it takes for each ball to reach that point.
For the ball thrown upward: The initial velocity (u) is +20 m/s (positive because it is upward). The acceleration due to gravity (g) is -9.8 m/s² (negative because it acts in the opposite direction to the initial velocity). The displacement (s) is the distance traveled by the ball until they meet.
We can use the equation of motion:
s = ut + (1/2)at²
Plugging in the known values:
s = 20t + (1/2)(-9.8)t²
For the ball dropped from the cliff: The initial velocity (u) is 0 m/s (as it is dropped from rest). The acceleration due to gravity (g) is -9.8 m/s² (negative because it acts downward). The displacement (s) is the distance traveled by the ball until they meet. It starts from a height of 50 meters.
We can use the equation of motion:
s = ut + (1/2)at²
Plugging in the known values:
s = 0t + (1/2)(-9.8)t² + 50
Now we can equate the two expressions for s since they represent the same location where the balls meet:
20t + (1/2)(-9.8)t² = (1/2)(-9.8)t² + 50
Simplifying the equation:
20t = 50
t = 50 / 20
t = 2.5 seconds
Therefore, the two balls meet after approximately 2.5 seconds.
To find the location where they meet, we can substitute this time value back into either of the equations. Let's use the equation for the ball thrown upward:
s = 20t + (1/2)(-9.8)t²
s = 20(2.5) + (1/2)(-9.8)(2.5)²
s = 50 + (-12.25)
s ≈ 37.75 meters
Therefore, the two balls meet approximately 37.75 meters above the ground.