Question

13. (40 pts) A baseball promoter mounts a baseball launcher on a car that launches baseballs at a muzzl velocity of 40 m/s. While the car is travelling forward at 15 mph the baseball launcher is shot at an angle 25° above horizontal in the direction of the car's motion. The ball leaves the launcher 7 ft above the ground. Assume no air resistance. A. Find the time, tr, when the baseball hits the ground after being launched. 8. Find the distance, xy, that the baseball has travelled in the horizontal direction when it hits the ground

Answer 1

Situation is From ground frame velocity of baseball in positive x - direction (v_x) = (40 cos (25)) + 15 times 1.609 times 5/18 = 36.252 + 6.704 = 42.956 m/s velocity of baseball in vertical direction (v_y) = 40 sin (25) = 16.904 m/s so it is a projectile motion height = 7 ft = 7 times 0.3048 = 2.134 meter time for maximum height t_f = Time of fligh + t prime = 24 sin theta/g + t prime for t prime apply S = u t prime + 1/2 at^2 in vertical direction 2.134 = 16.904 t prime + 1/2 times 9.8 t^12 = > 4.9 t^2 + 16.904 t - 2.134 = 0 t prime = -16.904 + squareroot (16.904)^2 + 4 times 4.9 times 2.134/2 times 4.9 = 1.195 sec \r\n t_f = 2 times 40 sin (25)/9.8 + 1.195 sec = 3.449 sec x_f = v_x t_f = 42.956 times 3.449