You drop a rock from rest out of a window on the top floor of a building, 50.0 m above the ground. When the rock has fallen 3.80 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the rock that your friend threw? I honestly just have no idea how to get that answer.
To determine the initial velocity of the rock thrown by your friend, we can use the principles of motion and apply the equations of kinematics. Specifically, we will use the equation for the position of an object in free fall:
y = v₀t + 0.5gt²
Where:
- y is the vertical displacement (distance fallen) of the object (3.80 m in this case)
- v₀ is the initial velocity of the object that your friend threw (what we're trying to find)
- t is the time taken for the object to fall (which will be the same for both rocks)
- g is the acceleration due to gravity (approximately 9.8 m/s²)
We can set up two separate equations using the data provided:
For the rock dropped by you:
y₁ = 0.5gt₁²
For the rock thrown by your friend:
y₂ = v₀t₂ + 0.5gt₂²
Given that both rocks reach the ground at the same time, we can assume that t₁ = t₂.
Now, let's solve for the initial velocity of the rock thrown by your friend (v₀):
y₂ = v₀t₂ + 0.5gt₂²
y₂ = (v₀t₁ + 0.5gt₁²) [Substituting t₁ = t₂]
y₂ = y₁ [Since both rocks reach the ground at the same time]
Substituting the known values:
3.80 m = 0 + 0.5(9.8 m/s²)t₁²
3.80 m = 4.9 m/s²t₁²
We can rearrange the equation to solve for t₁:
t₁² = (3.80 m) / 4.9 m/s²
Taking the square root of both sides:
t₁ ≈ √(3.80 m / 4.9 m/s²)
Now that we know the time it took for the rocks to reach the ground (t₁), we can substitute this value back into either of the initial equations to solve for v₀:
y₁ = 0.5gt₁²
3.80 m = 0.5(9.8 m/s²)(t₁)²
Solving for v₀:
v₀ = (3.80 m / 0.5(t₁)²) / (√(3.80 m / 4.9 m/s²))
Calculating the value will give you the initial velocity of the rock thrown by your friend.