Maurice Tutor

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Algebra,Applied Sciences,Biology,Calculus,Chemistry,Economics,English,Essay writing,Geography,Geology,Health & Medical,Physics,Science Hide all
Teaching Since: May 2017
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  • MCS,PHD
    Argosy University/ Phoniex University/
    Nov-2005 - Oct-2011

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  • Professor
    Phoniex University
    Oct-2001 - Nov-2016

Category > Management Posted 29 Jan 2018 My Price 10.00

Projectile-No Air Resistance

Range of a Projectile-No Air Resistance If you worked Problem 23 in Exercises 3.12, you saw that when air resistance and all other forces except its weight w = mg are ignored, the path of motion of a ballistic projectile, such as a cannon shell, is described by the system of linear differential equations

(a) If the projectile is launched from level ground with an initial velocity v0 assumed to be tangent to its path of motion or trajectory, then the initial conditions accompanying the system are   the initial speed is constant and θ is the constant angle of elevation of the cannon. See Figure 3.R.4 in The Paris Guns problem on page 204. Use the Laplace transform to solve system (8).

(b) The solutions x(t) and y(t) of the system in part (a) are parametric equations of the trajectory of the projectile. By using x(t) to eliminate the parameter t in y(t) show that the trajectory is parabolic.

(c) Use the results of part (b) to show that the horizontal range R of the projectile is given by

From (9) we not only see that R is a maximum when θ = π /4 but that a projectile launched at distinct complementary angles θ and π/2 - θ has the same submaximum range. See FIGURE 4.6.8. Use a trigonometric identity to prove this last result.

(d) Show that the maximum height Hof the projectile is given by

(e) Suppose g = 32 ft/s2, and θ = 38° and v0 = 300 ft/s. Use (9) and (10) to find the horizontal range and maximum height of the projectile. Repeat with θ = 52° and v0 = 300 ft/s.

(f) Because formulas (9) and (10) are not valid in all cases (see Problem 22), it will advantageous to you to remember that the range and maximum height of a ballistic projectile can be obtained by working directly with x(t) and y(t) that is, by solving y(t) = 0 and y' (t) = 0. The first equation gives the time when the projectile hits the ground and the second gives the time when y(t) is a maximum. Find these times and verify the range and maximum height obtained in part (e) for the trajectory with θ = 38° and v0 = 300 ft/s. Repeat with θ = 52°.

(g) With g = 32 ft/s2, θ = 38° and v0 = 300 ft/s use a graphing utility or CAS to plot the trajectory of the projectile defined by the parametric equations x(t) and y(t) in part (a). Repeat with θ = 52°. Using different colors superimpose both curves on the same coordinate system.

Problem 23

Projectile Motion A projectile shot from a gun has weight w = mg and velocity v tangent to its path of motion or trajectory. Ignoring air resistance and all other forces acting on the projectile except its weight, determine a system of differential equations that describes its path of motion. See FIGURE 3.12.4. Solve the system. [Hint: Use Newton's second law of motion in the x and y directions.]

 

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Status NEW Posted 29 Jan 2018 12:01 AM My Price 10.00

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