1. The problem statement, all variables and given/known data projectile intial velocity vo of 49 meters per second. angle theta intial is 76 degrees. projectile takes off and lands on the same horizontal plane. neglecting air resistance and using g=9.8 m/s squared. find a. range of projectile: 115.020m b. total time of flight: 9.70295 c. time in which the projectile is farthest from its launch point: 4.85147 d. farthest distance the projectile is from its launch point: 115.330 e. the critical angle theta c such that for theta c is greater than or equal to theta which is greater than 0 the projectile is always moving farther away from its launch point and for theta greater than theta c there are times during the projectiles trajectory when the projectile is moving towards its launch point: not sure where to begin give six significant digits in all answers 2. Relevant equations Vox = Vo Cos(theta); Voy = Vo sin(theta); 1/2t=Voy/ay; t=2(1/2t) dy=Voyt+.5ayt^2; dx=Voxt 3. The attempt at a solution Vox= 49 cos76= 11.85417 Voy= 49 sin76= 47.54449 1/2t= 47.54449/9.8= 4.851479 dy= (47.54449)(4.851479)+(.5)(-9.8)(4.851479)(4.851479)=230.6611-115.3306= 115.330 t= 2(4.851479)= 9.70295 dx= (11.85417)(9.702958)= 115.020 I have been out of school for quite sometime now and didn't do so well on physics then. I believe I am close on the range and total time. I feel I am missing something for farthest point. I was assuming this meant the peak of its arc from the ground. But after reading it again I am not so sure. And Theta C I dont get since its an arc not an angle unless its talking about the angle of launch. Any hint in the right direction would be appreciated. Thanks.