Answer:
To find the time at which the two balls will be at the same height above the ground, we need to compare their respective heights as functions of time.
Let's analyze the motion of each ball separately:
1. Ball thrown upward:
Using the equation for vertical motion, the height of the ball thrown upward can be represented as:
h₁ = h₀ + v₀t - (1/2)gt²,
where h₁ is the height, h₀ is the initial height (ground level), v₀ is the initial velocity, t is time, and g is the acceleration due to gravity (-9.8 m/s²).
For the ball thrown upward:
h₁ = 0 + (14.0 m/s)t - (1/2)(9.8 m/s²)t²,
h₁ = 14.0t - 4.9t².
2. Ball dropped from a building:
Since the ball is dropped, its initial velocity is 0. The height of the ball dropped from the building can be represented as:
h₂ = h₀ + v₀t - (1/2)gt²,
where h₂ is the height, h₀ is the initial height (14 m), v₀ is the initial velocity (0 m/s), t is time, and g is the acceleration due to gravity (-9.8 m/s²).
For the ball dropped from the building:
h₂ = 14.0 m + 0t - (1/2)(9.8 m/s²)t²,
h₂ = 14.0 - 4.9t².
To find the time at which the balls will be at the same height, we set h₁ equal to h₂ and solve for t:
14.0t - 4.9t² = 14.0 - 4.9t².
Simplifying the equation:
14.0t = 14.0,
t = 1.0 second.
Therefore, after 1.0 second, both balls will be at the same height above the ground.