a baseball leaves the bat with a speed of 44m/s and an angleof 30 degrees above the horizontal. a 5.0 m fence is located at ahorizontal distance of 132m from the point where the ball isstruck. assume the ball leaves the bat 1.0m above groundlevel. a. how long is the ball in the air from the bat to the time itgoes over the fence? b. by how much does the ball clear the fence? please show all work a baseball leaves the bat with a speed of 44m/s and an angleof 30 degrees above the horizontal. a 5.0 m fence is located at ahorizontal distance of 132m from the point where the ball isstruck. assume the ball leaves the bat 1.0m above groundlevel. a. how long is the ball in the air from the bat to the time itgoes over the fence? b. by how much does the ball clear the fence? please show all work
Explanation / Answer
Given that theinitial velocity is u = 44m/s The angle is = 300 Thehorizontal distance is D = 132m -------------------------------------------------------------------- (a) Horizontal distance covered by the ballis D = horizontal velocity*time of flight 132m= U cos*t t = 132m / Ucos =--------- sec (b) The verticle distance covered in time t fromheight 1.0m is From the equation of motion y = u sin*t +(1/2)at2 y = 44m/s*sin300*t - (1/2)gt2 =---------- m Then the separation between the fence and ball is d = y- 5.0m =------- m =---------- m Then the separation between the fence and ball is d = y- 5.0m =------- m