ID [THIS IS A MORE LENGTHY PROBLEM, SAVE IT FOR NEAR THE END] A special slingsho
ID: 1783406 • Letter: I
Question
ID [THIS IS A MORE LENGTHY PROBLEM, SAVE IT FOR NEAR THE END] A special slingshot is designed to launch an angry bird from the ground to a target away - also not kn test shot on the slingshot with a velocity of 12.0 m/s, but our test bird lands 1.0 meters beyond our target (it went too far). Calculate both the fixed launch angle of the slingshot and the necessary launch velocity to get our angry bird to land exactly at its target which is 13.7 meters on the ground. The angle of this special slingshot is unfortunately not adjustable, and also own; but the speed of the launch is adjustable and calibrated (adjustable and known). We fire a A) 30 degrees, 11.6 m/s B) 60 degrees, 11.6 m/s C) 60 degrees, 12.7 m/s ) 45 degrees, 12.7 m/s 45 degrees, 11.6 m/sExplanation / Answer
We know that, Range R = v^2 sin 2 theta /g
13.7+1 = 12^2 sin 2 theta /9.8
sin 2 theta = 9.8*14.7/12^2 = 1
2 theta = 90 degree
theta = 45 degree. answer
R = v^2 sin 2 theta /g
13.7 = v^2*1/9.8
v = sqrt(13.7*9.8) = 11.6 m/s answer
Hence the option E is correct