Question

Question 9 Your answer is parbally correct. Try again. A bal is shot from the gro und into the air. At a he ght ef 9.4 m, its velocity is V = 7.1 + 5 m/s, th horno eal and upward. (a) To what man r um height does the bal:ise? (b) What total hon or tal dstance does the ball travel? What are the (e) magnitude and (d) angle (below the horcontal; give as negative) of the balls velocity just before t hits the ground? (a) N (b) Numder (c) Number.981091248 (d) Number 137 76307797

Answer 1

(a) at the maximum height v'_y = 0 v = 7.1 i + 5.5 j using equation of the motion Here v'_y = 0, u_y = 5.5 m/s v^2_y = u^2_y - 2gh' 0 = (5.5)^2 - 2 times 9.8 times h' h' = (5.5)^2/2 times 9.8 = 1.543 m H_max = 9.4 + 1.543 H_max = 10.943 m (c) Time taken by the ball from P to 0 v^2_y = u^2_y + 2gH_ma Here v_y = at point O H_max = 10.943m u_y = 0 at point P y - component of the velocity v^2_y = 2 times 9.8 times 10.943 v_y = 14.645 m/s and x - component of the velocity v_x = 7.1 m/s since acceleration is only y - direction magnitude v = squareroot v^2_x + v^2_y = squareroot 14.645^2 + 7.1^2 v = 16.28 m/s (d) theta = tan^-1(-v_y/v_x) = tan^-1 (-14.645/7.1 = -64.14 degree) theta = -64.14 degree (-) show below the horizontal. If