A gun fires android shell at an angle of elevation 30 degree with a velocity of 2 *10m/s.what are the horizontal and vertical component of the velocity what is the range of the shell how high would the hall rise ？
To find the horizontal and vertical components of the velocity, we can use the trigonometric functions. Given that the angle of elevation is 30 degrees and the velocity is 2 * 10 m/s, we can calculate:
Horizontal Component of Velocity (Vx):
Vx = Velocity * cos(angle of elevation)
= 2 * 10 m/s * cos(30°)
Vertical Component of Velocity (Vy):
Vy = Velocity * sin(angle of elevation)
= 2 * 10 m/s * sin(30°)
Simplifying these calculations:
Vx = 2 * 10 m/s * (√3/2) = 10√3 m/s (approximately 17.32 m/s)
Vy = 2 * 10 m/s * (1/2) = 10 m/s
Now, to find the range of the shell, we can use the formula:
Range (R) = (Vx^2 * sin(2*angle of elevation)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values:
R = (10√3 m/s)^2 * sin(2*30°) / 9.8 m/s^2
Simplifying:
R = (300 * sin(60°)) / 9.8
R = (300 * √3/2) / 9.8
R = 150√3 / 9.8 (approximately 17.32)
Therefore, the range of the shell is approximately 17.32 meters.
To find the maximum height reached by the shell, we can use the formula:
Maximum Height (H) = (Vy^2) / (2 * g)
Substituting the known values:
H = (10 m/s)^2 / (2 * 9.8 m/s^2)
Simplifying:
H = 100 / 19.6
H = 5.1 meters (approximately)
So, the shell would rise to a height of approximately 5.1 meters.