Answer:
Maximum height reached is given as H = 5.00 m
Explanation:
As we know that if the ball is projected at some angle with the horizontal then the range of the projectile is given by the formula
![R = \frac{v^2 sin2\theta}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin2%5Ctheta%7D%7Bg%7D)
here we have
![R = 55 m](https://tex.z-dn.net/?f=R%20%3D%2055%20m)
![\theta = 20^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020%5Eo)
v = 40 m/s
Now we have
![55 = \frac{40^2 sin40}{g}](https://tex.z-dn.net/?f=55%20%3D%20%5Cfrac%7B40%5E2%20sin40%7D%7Bg%7D)
![g = 18.7 m/s^2](https://tex.z-dn.net/?f=g%20%3D%2018.7%20m%2Fs%5E2)
now for maximum height we have
![H = \frac{v^2sin^2\theta}{2g}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7Bv%5E2sin%5E2%5Ctheta%7D%7B2g%7D)
![H = \frac{40^2 sin^220}{2(18.7)}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B40%5E2%20sin%5E220%7D%7B2%2818.7%29%7D)
![H = 5 m](https://tex.z-dn.net/?f=H%20%3D%205%20m)
Answer:
![\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd_2%7D%7Bd_1%7D%20%3D%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B%5Csqrt%7B4%7D%7D%3D%5Csqrt%7B%5Cfrac%7B6%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B2%7D%3D%201.225)
Or in the other way:
![\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd_1%7D%7Bd_2%7D%20%3D%5Cfrac%7B%5Csqrt%7B4%7D%7D%7B%5Csqrt%7B6%7D%7D%3D%5Csqrt%7B%5Cfrac%7B4%7D%7B6%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B3%7D%3D%200.816)
Explanation:
For this case we need to remember that hectare is a measure of area. And we have the following ratios:
![r_1 =40000 \frac{plants}{acre}](https://tex.z-dn.net/?f=%20r_1%20%3D40000%20%5Cfrac%7Bplants%7D%7Bacre%7D)
We can convert this into ![plants/m^2](https://tex.z-dn.net/?f=plants%2Fm%5E2)
![r_1 = 40000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=4 \frac{plants}{m^2}](https://tex.z-dn.net/?f=%20r_1%20%3D%2040000%20%5Cfrac%7Bplants%7D%7Bacre%7D%20%2A%5Cfrac%7B1%20acre%7D%7B10000%20m%5E2%7D%3D4%20%5Cfrac%7Bplants%7D%7Bm%5E2%7D)
![r_2=60000 \frac{plants}{acre}](https://tex.z-dn.net/?f=%20r_2%3D60000%20%5Cfrac%7Bplants%7D%7Bacre%7D)
We can convert this into ![plants/m^2](https://tex.z-dn.net/?f=plants%2Fm%5E2)
![r_2 = 60000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=6 \frac{plants}{m^2}](https://tex.z-dn.net/?f=%20r_2%20%3D%2060000%20%5Cfrac%7Bplants%7D%7Bacre%7D%20%2A%5Cfrac%7B1%20acre%7D%7B10000%20m%5E2%7D%3D6%20%5Cfrac%7Bplants%7D%7Bm%5E2%7D)
By dimensional analysis we assume that
we can find the ratio in terms of distance like this:
![d_1 =\sqrt{4}= 2 \frac{\sqrt{plants}}{m}](https://tex.z-dn.net/?f=%20d_1%20%3D%5Csqrt%7B4%7D%3D%202%20%5Cfrac%7B%5Csqrt%7Bplants%7D%7D%7Bm%7D)
![d_2= \sqrt{6}= 2.449 \frac{\sqrt{plants}}{m}](https://tex.z-dn.net/?f=%20d_2%3D%20%5Csqrt%7B6%7D%3D%202.449%20%5Cfrac%7B%5Csqrt%7Bplants%7D%7D%7Bm%7D%20)
And if we find the ratios in terms of distance we got:
![\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd_2%7D%7Bd_1%7D%20%3D%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B%5Csqrt%7B4%7D%7D%3D%5Csqrt%7B%5Cfrac%7B6%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B2%7D%3D%201.225)
Or in the other way:
![\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd_1%7D%7Bd_2%7D%20%3D%5Cfrac%7B%5Csqrt%7B4%7D%7D%7B%5Csqrt%7B6%7D%7D%3D%5Csqrt%7B%5Cfrac%7B4%7D%7B6%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B3%7D%3D%200.816)
Answer:
When two or more forces are acting on a body, then the total of all the forces which causes the resulting effect is the resultant force or net force. As force is a vector, we need to take the vector sum of all the forces to calculate the resultant.
Explanation: