Question

A baseball player at second base throws a ball 87 feet to the player at first base. The ball is thrown 5 feet above the ground with an initial velocity of 50 miles per hour and at an angle of 16 above the horizontal. At which height does the player at first base catch the bal1? Round your answer to three decimal places. 15. a. 7.576 feet b. 6.676 feet c. 5.476 feet d. 6.076 feet e5.576 feet quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 38 yards directly downfield at a height of 4 feet. The pass is released at an angle of 45 with the horizontal. Find the speed of the football when it is released. Round your answer to three decimal places. 16. a. 33.169 feet per second b. 59.366 feet per second c. 19.462 feet per second d. 118.732 feet per second e. 190.165 feet per second The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by a receiver 76 yards directly downfield at a height of 2 feet. The pass is released at 17. an angle of 30 with the horizontal. Find the maximum height of the football. Round your answer to one decimal place. a. -377.3 feet b.-374.2 feet c. -425.8 feet d. -528.2 feet e. -583.0 feet The quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 46 yards directly downfield at a height of 4 feet. The pass is released at an angle of 30' with the horizontal. Find the time in seconds the receiver has to reach the proper position after the quarterback releases the football. Round your answer to one decimal place. 18. a. 7.0 seconds b. 2.3 seconds c. 4.1 seconds d. 0.7 second e. 1.1 seconds

Answer 1

1 yard = 3ft, g = 32 ft/s Let 'U' be the speed of the foot ball when it was released. Use the following equations: x (4 cos 45 degree) t: y = y_0 + (u sin 45 degree) t - 1/2 g t^2 for horizontal motion: 38 times 3 = (4 cos 45 degree) t rightarrow (1) for vertical motion. 4 = 8 + (u sin (45 degree) t - 1/2 g t^2 rightarrow (3) from equations (1) and (2) 0 = 4 + (u sin 45 degree) times (114/u cos 45 degree) - 1/2 times 32 (114/u cos 45 degree)^2 Implies 0 = 4 + 114 tan 45 degree - 16 (114/(u cos 45 degree)^2 Implies 0 = 118 - 16 ((114)^2/u^2 (0.70 times (cos 45 degree = 0.70710678) Implies 118 (u^2 (0.7071)^2 = 16 (114)^2 Implies u = squareroot 16 (114)^2/118 (0.7070678)^2 = 59.36658 Therefore u TildeEqual 59.366 ft/sec