A gun is shot horizontally from a building that is 10 meters high at a velocity of 220 meters/ second:
I need to know what Vx is and Vy along with answering Vx after 3 seconds Vy after 3 second.
We were given these equations
Vy= Vy0-9.8t
y= y0+Vy0t-4.9t^2
x=Vxt
,but I'm still not sure what I'm doing. I just need to know how to work these problems because I have 2 more after this!! Please Help!!
To solve this problem, you can use the given equations of motion in two dimensions.
In this case, the gun is shot horizontally, so the initial vertical velocity (Vy0) is 0. The initial horizontal velocity (Vx0) is given as 220 meters/second. The initial vertical position (y0) is 10 meters, and the initial horizontal position (x0) is not given, but we can assume it is 0.
1. Finding Vx after 3 seconds:
To find Vx after 3 seconds, you can use the equation x = Vxt, assuming there are no external forces affecting the horizontal motion.
x = Vx * t
Since the motion is horizontal, there is no acceleration (a = 0) in that direction.
Vx = x / t
Vx = 0 / 3
Vx = 0 meters/second
So, Vx after 3 seconds is 0 meters/second.
2. Finding Vy after 3 seconds:
To find Vy after 3 seconds, you can use the equation Vy = Vy0 - 9.8t, which gives you the vertical velocity at any given time.
Vy = Vy0 - 9.8t
Since the initial vertical velocity is 0 (Vy0 = 0), we have:
Vy = 0 - 9.8 * 3
Vy = -29.4 meters/second
So, Vy after 3 seconds is -29.4 meters/second, which means the object is moving downward with a velocity of 29.4 meters/second.
Note: Negative velocity indicates downward motion, while positive velocity indicates upward motion.