Dont bother to answer if your are going to coppy an answer . i need to understan

Question

Dont bother to answer if your are going to coppy an answer . i need to understand the problem not just copy it . Will rate right away for people who explain.

A fireworks shell is launched vertically from point A with speed sufficient to reach a maximum altitude of 500 ft. A steady horizontal wind causes a horizontal acceleration of 0.5 ft/s2without affecting the vertical motion. What angle a compensates for the wind such that the shell peaks directly over the launch point A? What is the maximum height h achievable when launched at this angle?

- the corect answer is alfa = 0.445 degrees and maximum heigh =500.

Dont bother to answer if your are going to coppy an answer . i need to understand the problem not just copy it . Will rate right away for people who explain.

Explanation / Answer

Since you want to understand the problem, I am not giving the total solution. :)

The key to this problem is to understand the physics involved. The initial velocity can be decomposed into two components. Vertical component = v cos(alpha) and horizontal component= v sin(alpha).

The vertical component is subjected to gravitational force, while the horizontal component is subjected to the acceleration due to wind

At a given time instant t after launching the shell, the components of velocities are: vertical = v cos(alpha) - gt and horizonal = v sin(alpha) - at.

Now, you know the maximum height. The time to reach maximum height = v x cos(alpha) x t - 0.5 gt^2. At this time, the net horizontal displacement = 0. i.e. v x sin(alpha) x t = 0.5 a x t^2.

I hope this clears your doubts

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