Respuesta :
Answer:
9 floors
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 1+\frac{1}{2}\times 9.81\times 1^2\\\Rightarrow s=4.905\ m[/tex]
So, one floor is 4.905 m tall
In 3 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 3+\frac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]
The distance the ball would have covered would be 44.145 m in 3 seconds
Number of floors
[tex]\frac{44.145}{4.905}=9[/tex]
So, the ball would have fallen 9 floors in 3 seconds
A ball released from rest from the twentieth floor of a building, after 3 s falls 9 floors, considering they are evenly spaced.
A ball is released from rest from the twentieth floor of a building. It experiences a uniformly accelerated rectilinear motion. We can find the displacement after 1 s using the following kinematic equation.
[tex]s = ut + \frac{1}{2} a t^{2} = (0m/s)(1s) + \frac{1}{2} (9.8m/s^{2} ) (1s)^{2} = 4.9 m[/tex]
where,
- s: displacement
- u: initial velocity
- a: acceleration (gravity = 9.8 m/s²)
- t: time
The height of each floor is 4.9 m.
We will use the same equation to find the displacement after 3 s.
[tex]s = ut + \frac{1}{2} a t^{2} = (0m/s)(1s) + \frac{1}{2} (9.8m/s^{2} ) (3s)^{2} = 44.1 m[/tex]
If the height of each floor is 4.9 m and it falls 44.1 m in 3 s, the number of floors it falls is:
[tex]44.1 m \times \frac{1Floor}{4.9m} = 9 Floor[/tex]
A ball released from rest from the twentieth floor of a building, after 3 s falls 9 floors, considering they are evenly spaced.
Learn more: https://brainly.com/question/13027863
