A bullet leaves a gun at an angle of 40 degrees to the horizontal. It reaches maximum height in 1.3 seconds. Determine the speed (velocity) with which the bullet left the gun (aka initial velocity of bullet).
1. t = (Vfy - Voy) / -9.8 m/s/s
The Attempt at a Solution
First, i found the component for initial vertical velocity by solving for Voy:
(knowing that vertical velocity at max height is zero, 0 - Voy = -Voy)
t = -Voy / -9.8 m/s/s = 1.3 sec., -9.8m/s/s times 1.3 sec. = -12.74 m/s component for initial vertical veloc. (Voy)
i understand that this is the time equation, but to solve for the initial vertical veloc component and hence use that to find Vo, i had to work backwards, solving for the unknowns with what equations i can use. i hope my reasoning was accurate.
Next, I used the equation(vertical veloc. component equation): Vosin40 = -12.74 m/s, then solved for Vo: (12.74 / sin40) = 19.81 m/s Initial Velocity when bullet leaves gun. This will be the same for both equations: Vocos40, Vosin40.
That is, 19.81 m/s = Vo in the equations Vocos40 and Vosin40
19.81cos40 (horizontal veloc. component) = 15.16 and
19.81sin40 (vert. veloc component) = 12.74
Vocos40 is not needed for this problem, but i just included to show that i understand what Vo means(initial velocity) as compared to say, Vox, Voy, etc.
Is this the correct way of finding the initial velocity? It seems to make sense to me but I'd like to make sure.