Answer:
The bucket was the dropped from 56 th floor.
Explanation:
Given that,
Height of floor = 4.9 m
Height of 14 floor = 68.6 m
Time taken = 1 sec
We need to calculate the speed of the bucket
Using equation of motion
![s=ut+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value into the formula
![68.6=v\times1+\dfrac{1}{2}\times9.8\times(1)^2](https://tex.z-dn.net/?f=68.6%3Dv%5Ctimes1%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%281%29%5E2)
![v=68.6-\dfrac{1}{2}\times9.8\times(1)^2](https://tex.z-dn.net/?f=v%3D68.6-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%281%29%5E2)
![v=63.7\ m/s](https://tex.z-dn.net/?f=v%3D63.7%5C%20m%2Fs)
We need to calculate the time
Using equation of motion
![v=u+gt](https://tex.z-dn.net/?f=v%3Du%2Bgt)
![t=\dfrac{v}{g}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bv%7D%7Bg%7D)
Put the value into the formula
![t=\dfrac{63.7}{9.8}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B63.7%7D%7B9.8%7D)
![t=6.5\ sec](https://tex.z-dn.net/?f=t%3D6.5%5C%20sec)
We need to calculate the distance
Using equation of motion
![s=ut+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![s=0+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3D0%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value into the formula
![s=\dfrac{1}{2}\times9.8\times(6.5)^2](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%286.5%29%5E2)
![s=207.025\ m](https://tex.z-dn.net/?f=s%3D207.025%5C%20m)
We need to calculate the number of floor
![n=\dfrac{s}{h_{f}}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7Bs%7D%7Bh_%7Bf%7D%7D)
Put the value into the formula
![n=\dfrac{207.025}{4.9}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B207.025%7D%7B4.9%7D)
![n=42.25\approx42](https://tex.z-dn.net/?f=n%3D42.25%5Capprox42)
The bucket was the dropped from
![f=14+42= 56](https://tex.z-dn.net/?f=f%3D14%2B42%3D%2056)
Hence, The bucket was the dropped from 56 th floor.
Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) ![x=\frac{1}{2}at^2+v_0t+x_0](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2%2Bv_0t%2Bx_0)
The equation for velocity v and a constant acceleration a is:
(ii) ![v=at+v_0](https://tex.z-dn.net/?f=v%3Dat%2Bv_0)
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) ![a = \frac{v -v_0}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%20-v_0%7D%7Bt%7D)
(iv) ![x = \frac{v-v_0}{2t}t^2+v_0t + x_0](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bv-v_0%7D%7B2t%7Dt%5E2%2Bv_0t%20%2B%20x_0)
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) ![x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7Bv_0%7D%7B2%7Dt%20%2B%20v_0t%3D%20%5Cfrac%7Bv_0%7D%7B2%7Dt)
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:![v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}](https://tex.z-dn.net/?f=v%3Dat%2Bv_0%3D0.2%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%2A%2010s%2B3%5Cfrac%7Bm%7D%7Bs%7D%3D2%5Cfrac%7Bm%7D%7Bs%7D%2B3%5Cfrac%7Bm%7D%7Bs%7D%3D5%5Cfrac%7Bm%7D%7Bs%7D)
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):
![f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bx%28t_1%29%7D%7Bx%28t_2%29%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Dat_1%5E2%20%7D%7B%5Cfrac%7B1%7D%7B2%7Dat_2%5E2%7D%3D%5Cfrac%7Bt_1%5E2%7D%7Bt_2%5E2%7D%3D%5Cfrac%7B10%5E2%7D%7B1%5E2%7D%3D%5Cfrac%7B100%7D%7B1%7D)
I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.
Since each light year is approximately 9 trillion kilometres, 4.80 light years is 43.2 trillion kilometres, or 43,200,000,000,000,000 metres