A new projectile launcher is developed in the year 2023 that can launch a 10,000

Question

A new projectile launcher is developed in the year 2023 that can launch a 10,0000kg spherical probe with initial speed of 6000 m/s. For testing purposes, objects are launched vertically.

a) Neglect air resistance and assume that acceleration of gravity is constant. Determine how high the launched object can reach above the surface of Earth.

b) If the object has radius of 20 cm and the air resistance is proportional to the square of the object's speed with C= 0.2, determine the maximum height reached. Assume the density of air is constant.

c)Now also include the fact the acceleration of gravity decreases as the object soars above Earth. Find the height reached.

d)Now add the effects of the decrease in air density by ln()=-0.05h + 0.11 where is the air density in kg/m^3 and h is the altitude above Earth in km. Determine how high the object now goes.

Explanation / Answer

a)To find this, you can use the equation (velocity at the top)^2 = (original velocity)^2 +2x acceleration due to gravity x height Then plug in that at the top, velocity is equivalent to 0 and solve for height. b) Now you will write a function for the velocity of said object by subtracting the frictional deceleration function from the typical acceleration constant. Then you will take the integral from initial velocity to final velocity. With that, you will be able to find height. c) Now you must write a function for the decrease in acceleration and integrate that with respect to 0 to height. d) You will now do the same thing as in part B, only your function will be representative of the changing air density as your height increases.

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