Question

tion 4 The following three questions refer to this problem: You drop a 2.1 kg ball from a height of 2.2 m and notice that it bounces back to a height of 1.9 m. Somehow, you measure that the ball is in contact wi 4 out of 4 po the ground for 96 ms. Note that the term "drop" implies the initial velocity is zero What was the speed of the ball the instant before it hits the ground? Selected Answer: 6.57 Correct Answer 6.6+02 Question 5 0 out of 4 points What was the speed of the ball the instant it leaves the ground after bouncing? Hint. The speed you are looking for is the speed the ball needs to reach a height of 1.9 m. This is an equivalent statement to asking: What would the speed of the ball be if it was dropped from a height of 1.9 m. Remember that the up and "down" part of free-fall motion are symmetric if the objocts starts and stops at the same height (i.e., vertical displacement is zero) Selected Answer 5.63 Correct Answer 6.1+0.2

Answer 1

Electric forces at corner of square F_14 = kq_1 q_4/r_14^2 = 9 times 10^9 times 3 times 10^-6 times 2.4 times 10^-6/(0.15)^2 F_14 = 2.88 N and F_34 = kq_3 q_4/r_34^2 = 9 times 10^9 times 6 times 10^-6 7 3 times 10^-6/(0.15)^2 F_34 = 7.2 N F_42 = Kq_2 q_4/r_24^2 = 9 times 10^9 times 9 times 10^-6 times 3 times 10^-6/(15 squareroot 2 times 10^-2)^2 F_42 = 5.4 N (a) Then Force on q_4 = 3 mu c is Fx = F_34 + F_42 cos 45 degree Fx = 7.2 + 5.4/squareroot 2 = 11 N and Fy = -F_14 - F_42 sin 45 degree = -6.7 N then F_Total = 12.88 N and angle will be theta = tan^-1(6.7/11) = 31 degree below + x-axis (b) F_13 = kq_1q_3/r^2 = 9 times 10^9 times 2.4 times 10^-6 times 6 times 10^-6/(.15 squareroot 2) F_13 = 2.88 N and F_23 = kq_2 q_3/r_23^2 = 9 times 10^9