- An object that is thrown vertically upward will return to its original position with the same speed as it had initially if air resistance is negligible. If air resistance is appreciable, will this result be altered, and if so, how? [Hint: The acceleration due to air resistance is always in a direction opposite to the motion.]
Slower speed when falling down.
The result will be altered so that the speed at any given height is less than it would have been without air resistance. Air resistance is a force dependent on the speed of the object. The object is initially thrown with a kinetic energy 0.5mv^2. Using the work equation w = fd then you can same a model of the effect that air resistance would have on the motion. Since air resistance is a function of speed which decreases as speed decreases you could put any equation f(v) which has a positive slope to model the force from air resistance. Air resistance and gravity are both doing work on the object while it is moving with the work from gravity being w_g = mgd. Combining this work with the initial kinetic energy of the object gives KE = 0.5m(v initial)^2 - mgd - f(v)d. Removing the -f(v)d term gives the equation for negligible air resistance. Plugging this energy equation back into the kinetic energy equation you get 0.5m(v initial)^2 - mgd - f(v)d = 0.5mv^2. Isolating for v gives [(v initial)^2 - 2gd - 2f(v)d/m]^0.5 = v. From inspection of the equation you can see that the term for the work of air resistance, -2f(v)d/m reduces the velocity at any given distance, d. The larger the air resistance, f(v), the lower the velocity will be at any distance, d.