A beanbag was thrown straight upward from the ground at a initial velocity of 22.5 m/s. At the same time, another beanbag is dropped from the top of 17. 0 m building. A what time will the two beanbags be at the same height above the ground?

going upward:

hf=22.5*t-4.9t^2
going downward:
hf=17-4.9t^2

but hf=hf, and times are same, so
17-4.9t^2=22.5*t-4.9t^2
or t=17/22.5 seconds
hf=17-4.9(17/22.5)^2=14.2 ? meters

To find the time at which the two beanbags will be at the same height above the ground, we need to determine the height of each beanbag as a function of time.

Let's start with the beanbag thrown upward. We can use the equation of motion for vertical motion:

h₁ = v₁₀t + (1/2)at²

Where:
h₁ is the height above the ground,
v₁₀ is the initial velocity (22.5 m/s),
t is the time,
a is the acceleration (considered negative due to gravity, -9.8 m/s²).

For the beanbag dropped from the building, we can use the same equation, but with a different initial velocity since it was dropped instead of thrown:

h₂ = v₂₀t + (1/2)at²

Where:
h₂ is the height above the ground for the dropped beanbag,
v₂₀ is the initial velocity (0 m/s),
t is the time,
a is the acceleration (-9.8 m/s²).

Since we want to find the time at which both beanbags are at the same height, we can set the two height equations equal to each other:

h₁ = h₂

v₁₀t + (1/2)at² = v₂₀t + (1/2)at²

Now we can substitute the values into the equation:

22.5t + (1/2)(-9.8)t² = (1/2)(-9.8)t²

The t² terms cancel out:

22.5t = 0

Divide both sides by 22.5:

t = 0

Therefore, both beanbags will be at the same height above the ground at t = 0 seconds, which means they were at the same height at the initial time.