The height of the roof is <u>3.57m</u>
Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.
Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.
![s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2](https://tex.z-dn.net/?f=s_2%3D%5Cfrac%7B1%7D%7B2%7D%20g%284t%29%5E2%3D8gt%5E2%5C%5C%20%20s_3%3D%5Cfrac%7B1%7D%7B2%7D%20g%283t%29%5E2%3D%284.5%29gt%5E2)
The length of the window s is given by,
![s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2](https://tex.z-dn.net/?f=s%3Ds_2-s_3%5C%5C%20%281.00%20m%29%3D8gt%5E2-4.5gt%5E2%3D3.5gt%5E2%5C%5C%20t%5E2%3D%5Cfrac%7B1.00%20m%7D%7B%283.5%29%289.8m%2Fs%5E2%29%7D%20%3D0.02915s%5E2)
The first drop is at the bottom and it takes 5t seconds to reach down.
The height of the roof h is the distance traveled by the first drop and is given by,
![h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B1%7D%7B2%7D%20g%285t%29%5E2%3D%5Cfrac%7B25t%5E2%7D%7B2g%7D%20%3D%5Cfrac%7B25%280.02915s%5E2%29%7D%7B2%289.8m%2Fs%5E2%29%7D%20%3D3.57%20m)
the height of the roof is 3.57 m
The answer is 330 meters per second! hope this helped! please crown me if it did when you can!
Answer:
u= 35.30 s
Explanation:
given,
horizontal distance covered by the ball = 105 m
vertical distance to clear by the ball = 4 m
angle at which the ball was hit = 31°
Let the initial velocity is u m/s and the ball take t sec to reach the fence.
![R = u_x t](https://tex.z-dn.net/?f=R%20%3D%20u_x%20t)
![R = u cos \theta \times t](https://tex.z-dn.net/?f=R%20%3D%20u%20cos%20%5Ctheta%20%5Ctimes%20t)
![105 = ut \times cos 31^0](https://tex.z-dn.net/?f=105%20%3D%20ut%20%5Ctimes%20cos%2031%5E0)
ut = 122.5
Using
![s = ut + \dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%20%3D%20ut%20%2B%20%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![4 = u (sin \theta) t - \dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=4%20%3D%20u%20%28sin%20%5Ctheta%29%20t%20-%20%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![4 = u (sin 31^0) t - \dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=4%20%3D%20u%20%28sin%2031%5E0%29%20t%20-%20%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![4 = 122.5 \times 0.52 - 0.5\times 9.8 \times t^2](https://tex.z-dn.net/?f=4%20%3D%20122.5%20%5Ctimes%200.52%20-%200.5%5Ctimes%209.8%20%5Ctimes%20t%5E2)
![t^2 = 12.06](https://tex.z-dn.net/?f=t%5E2%20%3D%2012.06)
![t = \sqrt{12.06}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B12.06%7D)
t = 3.47 s
now,
![u = \dfrac{122.5}{3.47}](https://tex.z-dn.net/?f=u%20%3D%20%5Cdfrac%7B122.5%7D%7B3.47%7D)
u= 35.30 s
Speed of the ball when it leaves the bat u= 35.30 s
Explanation:
time is nature's expression of change.
every change takes time.
it takes time to e.g. move from A to B.
at one point of time the universe is in one state, then some change happens, and the universe is in another state, and we have another point of time.
without time there is no change, without change there is no time.
Answer:
Polar substances are likely to dissolve in polar solvents. For example, ionic compounds, which are very polar, are often soluble in the polar solvent water. Nonpolar substances are likely to dissolve in nonpolar solvents.