show that time of ascent=time of descent

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show that time of ascent=time of descent

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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when the velocity of a body increses air resistance increases and so the weigt decreases
set the derivative of vertical location with respect to time to zero and solve for time. This gives you the time when speed is zero, which is when the object stops ascending and begins descending. Compare that result with vertical location = 0, which has two solutions: 0, when the object is released and the time when the object returns to the ground. Example: Y-Y0 = V0*sinA*time_bottom - (g/2)*time_bottom^2 = 0 Add the acceleration term to get: V0*sinA*time_bottom = (g/2)*time_bottom^2 Divide by t (t is not zero, because you don't care for that solution, so division is allowed) V0*sinA = (g/2)*time_bottom divide by (g/2) to get an expression for t time_bottom = 2*V0*sinA/g now take the derivative of the initial with respect to time and set to zero (velocity in the vertical direction is zero when the object is at it's peak) V0*sinA - 2*(g/2)*time_top = 0 a little manipulation yields: V0*sinA/g = time_top Combine the two equations to show that: time_top * 2 = time_bottom
Let the initial velocity be u During ascent: acceleration = -g Ascent completes when its velocity becomes zero. (=> final velocity v = 0) Using v=u+at, 0 = u + (-g)t = u -gt Therefore, t (time of ascent) = u/g And height to which the object ascends is given by s = ut + (1/2)at^2 = u(u/g) + (1/2)(-g)(u*u)/g*g = u*u/g - u*u/2g = u*u/2g During descent, initial velocity = 0, the distance it has to descend = u*u/2g, and the acceleration = g again using s = ut + (1/2)(at*t), u*u/2g = 0 + (1/2)(g*t*t) => g*t*t = u*u/g => t (time of descent) = u/g Therefore time of ascent = time of descent

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