Answer:
The 3rd image
Explanation:
I chose that because that's what I learned in biology
Answer:
(a)
vf_1s = 5.19 m/s
h_1s = 10.095 m
vf_4s = 24.23 m/s
h_4s = 4.49 m (below railing)
(b)
vi = 9.9 m/s
(c)
t = 1.53 s
h = 34.41 m
Explanation:
(c)
First, we will use the 1st equation of motion to find the time to attain the highest point:
![v_{f} = v_{i} + gt\\t = \frac{v_{i} - v_{f}}{g}\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Ct%20%3D%20%5Cfrac%7Bv_%7Bi%7D%20-%20v_%7Bf%7D%7D%7Bg%7D%5C%5C)
where,
t = time to attain maximum height = ?
vf = final velocity = 0 m/s (ball momentarily stops at highest point
vi = initial velocity = 15 m/s
g = - 9.81 m/s (for upward motion)
![t = \frac{0\ m/s - 15\ m/s}{- 9.81\ m/s^2}\\](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0%5C%20m%2Fs%20-%2015%5C%20m%2Fs%7D%7B-%209.81%5C%20m%2Fs%5E2%7D%5C%5C)
<u>t = 1.53 s</u>
Now, for the height attained we will use the 2nd equation of motion:
![h = v_{i}t + \frac{1}{2}gt^2\\\\h = (15\ m/s)(1.53\ s) + \frac{1}{2}(9.81\ m/s^2)(1.53)^2\\\\h = 22.93\ m + 11.48\ m](https://tex.z-dn.net/?f=h%20%3D%20v_%7Bi%7Dt%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ch%20%3D%20%2815%5C%20m%2Fs%29%281.53%5C%20s%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%289.81%5C%20m%2Fs%5E2%29%281.53%29%5E2%5C%5C%5C%5Ch%20%3D%2022.93%5C%20m%20%2B%2011.48%5C%20m)
<u>h = 34.41 m</u>
(b)
using the 3rd equation of motion for a height of 5 m:
![2gh = v_{f}^2 - v_{i}^2\\2(-9.81\ m/s^2)(5\ m) = (0\ m/s)^2 - v_{i}^2 \\v_{i} = \sqrt{98.1\ m^2/ s^2}\\](https://tex.z-dn.net/?f=2gh%20%3D%20v_%7Bf%7D%5E2%20-%20v_%7Bi%7D%5E2%5C%5C2%28-9.81%5C%20m%2Fs%5E2%29%285%5C%20m%29%20%3D%20%280%5C%20m%2Fs%29%5E2%20-%20v_%7Bi%7D%5E2%20%5C%5Cv_%7Bi%7D%20%3D%20%5Csqrt%7B98.1%5C%20m%5E2%2F%20s%5E2%7D%5C%5C)
<u>vi = 9.9 m/s</u>
<u></u>
(c)
At t = 1 s:
![v_{f1s} = v_{i} + gt\\v_{f1s} = 15\ m/s + (-9.81\ m/s^2)(1\ s)\\](https://tex.z-dn.net/?f=v_%7Bf1s%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Cv_%7Bf1s%7D%20%3D%2015%5C%20m%2Fs%20%2B%20%28-9.81%5C%20m%2Fs%5E2%29%281%5C%20s%29%5C%5C)
<u>vf_1s = 5.19 m/s</u>
<u></u>
<u></u>
<u>h_1s = 10.095 m</u>
<u></u>
At t = 4 s:
Since the ball covers the maximum height of 34.41 m in 1.53 s and then starts moving downward.
Therefore for the remaining 4 s - 1.53 s = 2.47 s, the initial velocity will be 0 m/s at the highest point and the value of g will be positive due to downward motion.
![v_{f4s} = v_{i} + gt\\v_{f4s} = 0\ m/s + (9.81\ m/s^2)(2.47\ s)\\](https://tex.z-dn.net/?f=v_%7Bf4s%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Cv_%7Bf4s%7D%20%3D%200%5C%20m%2Fs%20%2B%20%289.81%5C%20m%2Fs%5E2%29%282.47%5C%20s%29%5C%5C)
<u>vf_4s = 24.23 m/s</u>
<u></u>
<u></u>
now, for the position with respect to railing:
![h_{4s} = h - h_{down}\\h_{4s} = 34.41\ m - 29.92\ m\\](https://tex.z-dn.net/?f=h_%7B4s%7D%20%3D%20h%20-%20h_%7Bdown%7D%5C%5Ch_%7B4s%7D%20%3D%2034.41%5C%20m%20-%2029.92%5C%20m%5C%5C)
<u>h_4s = 4.49 m (below railing)</u>
<u></u>
Answer:
Part a)
![\theta = tan^{-1}\frac{a}{g}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7Ba%7D%7Bg%7D)
so here the angle made by the string is independent of the mass
Part b)
![a = 4.16 m/s^2](https://tex.z-dn.net/?f=a%20%3D%204.16%20m%2Fs%5E2)
Explanation:
Part a)
Let the string makes some angle with the vertical so we have force equation given as
![Tcos\theta = mg](https://tex.z-dn.net/?f=Tcos%5Ctheta%20%3D%20mg)
![T sin\theta = ma](https://tex.z-dn.net/?f=T%20sin%5Ctheta%20%3D%20ma)
so we will have
![tan\theta = \frac{ma}{mg}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7Bma%7D%7Bmg%7D)
![\theta = tan^{-1}\frac{a}{g}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7Ba%7D%7Bg%7D)
so here the angle made by the string is independent of the mass
Part b)
Now from above equation if we know that angle made by the string is
![\theta = 23 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2023%20degree)
so we will have
![tan23 = \frac{a}{g}](https://tex.z-dn.net/?f=tan23%20%3D%20%5Cfrac%7Ba%7D%7Bg%7D)
![a = g tan23](https://tex.z-dn.net/?f=a%20%3D%20g%20tan23)
![a = 9.81(tan23)](https://tex.z-dn.net/?f=a%20%3D%209.81%28tan23%29)
![a = 4.16 m/s^2](https://tex.z-dn.net/?f=a%20%3D%204.16%20m%2Fs%5E2)
The average distance between the Earth and the Sun is 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km<span> (584 million mi).</span>