Question

*Problem 2.35 A ball is kicked from a location 9, Ο 4> m on the ground with initial velocityく10 13 0 (a) What is the velocity of the ball 0.2 s after being kicked? (Use the Momentum Principle!) 6> m/s. The ball's speed is low enough that air resistance is negligible. m/s (b) In this situation (constant force), which velocity will give the most accurate value for the location of the ball 0.2 s after it is kicked? C C the initial velocity of the ball the arithmetic average of the initial and final velocities the final velocity of the ball (c) What is the average velocity of the ball over this time interval (a vector)? (d) Use the average velocity to find the location of the ball 0.2 s after being kicked 10, Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes (e) What is the y component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)? 10 m/s (f) Fill in the missing numbers in the equation below (update form of the Momentum Principle: mvy mVyi Fnet,yAr (g) How long does it take for the ball to reach its highest point? (h) Knowing this time, first find the y component of the average velocity during this time interval, then use it to find the maximum height attained by the bal -14 m

Answer 1

r_1 = 9i^^-4k^^u = -10i^^+ 13.0j^^-6k^^m/s a) t = 0.2 s When ball is kicked the only velocity along y direction will change So g = 9.8 m/s^2 Using equation of motion v = u - gt v = 13.0 - 9.8 times 0.2 v = 11.04 m/s So new velocity will be v = (-10.1104, -6) m/s b) The accurate velocity given by The arithmetic average of the initial and final velocities c) Average velocity v_avg = (-10, 13.0 + 11.04/2, -6) m/s v_avg = (10, 12.06, -6) m/s d) Location of the object d = (9 - 10 times 002, 0 + 12.02 times 0.2, -4 - 6 times 0.2) d = (7, 2.404, -5.2) m e) When ball reaches highest point v_yf = 0 m/s f) m times 0 = m times 13 + (-mg) Delta t g) Delta t = u/g = 13.0 m/s/9.8 m/s