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
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
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.