#### Transcript Calculus Gravity 3

PROJECT REAL WORLD: WHICH IS FASTER, GOING UP OR COMING DOWN?? By Moccia, Ariel, Stephanie, Taryn, and Giselle PROBLEM: A ball with a mass (m) is projected upward from Earth’s surface with a positive initial velocity (V0). We assume the forces acting on the ball are the force of gravity and a retarding force of air resistances with direction opposite to the direction of motion and with magnitude p|v(t)|, where p is a positive constant and v(t) is the velocity of the ball at time t. In both the ascent and the descent, the total force acting on the ball is –pv-mg. Confusing right? Well this is what we found from this complicated problem •Air resistance and gravity are both acting upon the ball (not in a vacuum). •V(t)=velocity •M=mass •T=Time •G=Acceleration due to gravity •P=Positive constant (air resistance) •-pv-mg= The total force acting on the ball, ascending and descending Part One: Prove that V(t) = (v0 + mg/p)ept/m -mg/p is the differential equation of mv = -pv-mg Steps: *v=dv/dt 1. Separate unlike terms, so that dv is with v and dt is alone. 2. Now we are going to find the integral of each. 3. Take out the m. 4. Notice what U is (the inner function), and what du/dv is (the derivative of the function) 5. Multiply in a ‘-p’ so you can use the substitution method, do not to put divide by a -p outside of the integral. 6. Use the substitution method now, and simplify. 7. 8. 9. 10. 11. 12. 13. Find the antiderivative of the integrals now. Isolate the ln(-pv-mg) by multiplying the other side by the reciprocal of –m/p. (pt/m)(C) is just another constant, D. There is a rule for ln that says if ln(a)=b, then a= eb. So, -pv-mg= e-pt/m + D. We can separate e-pt/m + D into e-pt/m eD. eD is just another constant, so we can call it K. Thus, we get –pv-mg =Ke-pt/m. Now we have to find what K, the constant, is equal to. The time (t) is 0, so we can plug that in, e0 = 1, so we have isolated K. Plug that back into the equation where K is. Isolate V, and after you can see that V(t)= (V0+ mg/p)ept/m –mg/p. Part Two: Show that the height of the ball, until it hits the ground is y(t)= (V0 + mg/p) m/p (1 – e-pt/m) –mgt/p Steps: 1. The derivative of the velocity equation will give us the height. 2. Find the integral of dy and the integral of the other side of the equation. 3. The derivative of e-pt/m is e-pt/m/-p/m. This is the same thing as e-pt/m × (-m/p). 4. Mg/p now become mgt/p, because the antiderivative of t0 is t1/1. 5. Even though you have Y now, you still have a constant, so solve for the constant (C). 6. Plug C back into the equation. 7. Take out like terms. 8. Now you have the equation for height. Part Three: If t1 is the time that the ball takes to reach its maximum height, show that t1= m/p ln[(mg+pv0)/mg]. Then find the time for a ball with a mass of 1 kg and an initial velocity of 20 m/s, and with air resistance being 1/10 of the speed. Steps: 1. Set the equation of v(t) equal to zero, because at the maximum height the ball stops for a second (thus the velocity is zero). This is so we can solve for t1. 2. Isolate the part that has t in it, so make sure that e-pt/m is isolated. 3. Using lna = B, a=eb, we see –pt/m is equal to ln( mg/pv0 + mg). 4. Use the rule rloga = Logar to solve for t. 5. 6. V1 = 20 m/s, p = 1/10, m= 1 kg. Plug in all the given values to find t1. Part Four: Let t2 be the time at which the ball falls back to Earth. For the same ball in Problem 3, estimate t2 by using the graph of the height function (y). Which is faster, going up or coming down? Steps: 1. Double t1. 2. If it goes the same speed going up as it does coming down, then the time will be the same. Thus 2t will be the total time. 3. Plug in 2t (3.71) into the y(t) equation, and solve for y. The y will tell you how far from reaching the ground the ball is. CONCLUSION? The ball is faster going up than it is coming down! Thanks Calculus!!