A projectile is launched at an angle of 55.6 degrees to the horizontal and follows a straight path with a speed of 7550 m/min. Find the vertical and horizontale components of this velocity.
Vo = 7550m/min @ 55.6deg.
Xo = hor. = 7550cos55.6 = 4265.5m/min.
Yo = ver. = 7550sin55.6 = 6229.6m/min.
69.2
To find the vertical and horizontal components of the velocity, we can use trigonometry.
The given angle of launch is 55.6 degrees. Let's call the vertical component of velocity Vv and the horizontal component of velocity Vh.
To find Vv, we can use the sine function:
Vv = V * sin(angle)
Substituting the given values, we have:
Vv = 7550 m/min * sin(55.6 degrees)
To find Vh, we can use the cosine function:
Vh = V * cos(angle)
Substituting the given values, we have:
Vh = 7550 m/min * cos(55.6 degrees)
Hence, the calculations are:
Vv = 7550 * sin(55.6)
Vh = 7550 * cos(55.6)
Calculating these values:
Vv ≈ 6382.9 m/min
Vh ≈ 5290.2 m/min
So, the vertical component of velocity is approximately 6382.9 m/min, and the horizontal component of velocity is approximately 5290.2 m/min.