A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.
Consider your advice to an artillery officer who has the following problem. From his current postition, he must shoot over a hill of height H at a target on the other side, which has the same elevation as his gun. He knows from his accurate map both the bearing and the distance R to the target and also that the hill is halfway to the target. To shoot as accurately as possible, he wants the projectile to just barely pass above the hill.
What is the initial speed?
Express v_0 in terms of g, R, and H

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT =  (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) =  (sinθ)√(2gH

v₀ = √(gR/(sin2θ))

how to make a simple air condition

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.