That means “ if possible then link”
Answer:
Δx = 25 ft.
Explanation:
Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:
1 ) Egg falling:
If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:
yf-y₀ = 1/2* g* t²
If we take the up direction as positive, we can solve for t as follows:
0-100 ft = 1/2* (-32.15 ft/s²)* t²
⇒ t = 
= 2.5 sec.
2) Person on the ground running away:
In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.
We can get the distance at which he can reach, applying the definition of velocity:
v = (xf-x₀) / (tfi-t₀)
If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:
xf = v*t = 20 ft/sec*1.25 sec = 25 ft.
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
the reason paper airplanes fly is because of lift the foil has no lift
Explanation:
