Calculating Vertical Height & Speed of Bullet Fired Upwards

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To calculate the vertical height and speed of a bullet fired upwards at 70 m/s with air resistance, the equation of motion must account for both gravity and air resistance, which is proportional to speed. The relevant equation is m*g - k*v = m*dV/dT, where gravity and air resistance act in opposite directions when the bullet ascends. The peak height and time to reach it can be determined by integrating this equation, while the speed upon return to the starting point will equal the initial speed, and the time taken to descend can also be calculated. The acceleration remains constant at g, but the effective acceleration during ascent will be reduced by the air resistance. Understanding these principles is crucial for solving the problem accurately.
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A bullet fired vertically upwards with a beginning speed 70 m / s. In this task, we shall assume that the sphere is exposed to air resistance is proportional to the speed, and given by. Proportionality constant k = 1.6 * 10 ^ -2 kg / s and the bullets mass m = 0.100 kg. We expect that the acceleration due to gravity is constant g = 9.81 m/s2.

a) Calculate how high the ball is coming, and how long it takes to peak.

b) What speed is the bullet when it comes back to the starting point, and how long does it take from the vertex to the starting point? What is the acceleration, then?

I guess I have to use integration on the air resistance, but I don't know how to start on this question. Anyone got a sugestion?
 
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Perhaps you start with Newton's second law?

ehild
 
I know about Newtons law..

In free fall, with air resistance, the equation is something like:

m*g - k*v = m*dV/dT.

But this is for free fall, I guess the one I need is a little bit different?
 
It is almost the same, but take care of the signs. If the bullet goes upward, both gravity and air resistance point downward.

ehild
 

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