02 May 5. Problems 2.5 The height of a ball that is thrown straight up with a certain force is a function of the time (t) from which it released given by At)–0.5 g40r (where g is a positive constant determined by gravity) The time at which the ball is at its maximum height depends on the parameter g, and is given by * Substituting t* back into the function f(t) yields f(t*) – dfíi*) ,the derivative (with respect to g) of which is equal to This shows how the maximum height of the ball changes as the parameter g changes Holding t constant at t*, the time at which the ball reaches its maximum height, yields the function f t*) =-0 5gtg2 + 40t*. Taking the derivative of this function with respect to g yields noyields . After plugging in the formula for t into result for aft*) than what you previously calculated. If g 32, as it does on Earth, the height at which the ball reaches its maximum height is t* ball is and the maximum height reached by the If g increases by a small amount, up to g 32.1, then the maximum height reached by the ball changes to approximately This result been predicted by the envelope theorem
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