
#1
Apr512, 08:13 PM

P: 7

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! 



#2
Apr512, 08:23 PM

P: 961

Relevant Equation
[itex]Y=(V_{iy})t\frac{1}{2}at^2[/itex] 



#3
Apr512, 08:27 PM

P: 280

The question is unanswerable, because you are not given knowledge about the initial ycomponent 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. 


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