Two students are on a balcony 22 m above the street. One student throws a ball vertically downward at 15.5 m/s; at the same instant, the other student throws a ball vertically up- ward at the same speed. The second ball just misses the balcony on the way down.
How far apart are the balls 0.357 s after they are thrown? The acceleration of gravity is 9.8 m/s2 .
x1 = 22 - 15.5(.357) - 4.9(.357^2)
x2 = 22 + 15.5(.357) - 4.9(.357^2)
Find the diff.
To find the distance between the two balls after 0.357s, we need to determine the positions of the two balls at that time.
Let's start by finding the position of the first ball that was thrown downward. We will use the equation for vertical displacement with constant acceleration:
d = v₀t + (1/2)at²
Where:
d = displacement (position)
v₀ = initial velocity
t = time
a = acceleration
For the downward thrown ball:
v₀ = 15.5 m/s (downward)
t = 0.357 s
a = 9.8 m/s² (downward)
Using the equation, we can calculate the vertical displacement of the first ball after 0.357s.
d₁ = (15.5 m/s)(0.357 s) + (1/2)(9.8 m/s²)(0.357 s)²
Next, let's find the position of the second ball thrown upward. Since we know it missed the balcony on its way down, we can assume it reached the same maximum height as the balcony before coming back down. The position of the second ball would be the height of the balcony plus its displacement above the balcony when they were both thrown.
For the upward thrown ball:
v₀ = 15.5 m/s (upward)
t = 0.357 s
a = -9.8 m/s² (downward)
Using the same equation, we can calculate the vertical displacement of the second ball after 0.357s:
d₂ = (15.5 m/s)(0.357 s) + (1/2)(-9.8 m/s²)(0.357 s)²
Now, we can find the distance between the two balls 0.357s after they were thrown by subtracting the positions:
distance = d₂ - d₁
Plug in the previously calculated values to get the final answer.