A bullet is fired at an angle of 60° with an initial velocity of 200 m/s. How long is the bullet in the air? 2. Relevant equations Vix = Vicosθ Viy = Visinθ Vfy = Viy + ayΔt Xf = xi + VixΔt ... Ask for more equations available.... 3. The attempt at a solution Vix = 200cos60° = 100m/s Viy = 200sin60° = 173.2 m/s X variables: xi = 0m xf = ? Δt =? Vix = 100m/s Y variables: Yi = ? Yf = 0m Δt =? Viy = 173.2m/s Vfy = ? a = -9.8m/s ANSWER HAS TO EQUAL 35.4 SECONDS, ACCORDING TO MY PHYSICS TEACHER!
The question is unanswerable, because you are not given knowledge about the initial y-component position of the bullet or the landscape into which the bullet will travel. But if we assume your teacher meant to describe a world in which bullets are fired from ground level into landscapes that are perfectly flat, we can proceed. If that is the case, Yi = 0. If you need additional help, just ask for it.