Suppose you throw a ball with horizontal part of the velocity x'(0) = 20 m/s and vertical part of the velocity y'(0) = 10 m/s. assuming that the gravitational force is constant (g = 9.8 m/s2) and that air resistance is proportional to the square of the velocity we obtain z" =-kk' /(x)2 + (y')2 " =-9 - ky' (x)2 + (y')2 x(0) = 0, x'(0) = 20, y(0) = 0, y(0) = 10 assuming the air resistance constant k= 0.1 m-1 find (a) the maximum height the ball will attain (b) how far from you will the ball hit the ground. details: for solutions of all differential equations you need to use the classical runge-kutta method developed in class.

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answer: howdy

step-by-step explanation:

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

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step-by-step explanation: