A gun fires a shell at an angle of elevation of 30* with a velocity of 2000ms-1.(A)what is the range of the shell.
time in air:
hf=hi+2000sin30*t-4.9t^2
hf=hi=0
t=2000sin30/4.9
range:
d=2000*cos30*timeinair
the fine article in wikipedia on "trajectory" explains this and more. The formula for the range is
R = v^2/g sin(2θ)
To find the range of the shell fired at an angle of elevation, we can use the equations of projectile motion. The range, R, can be calculated using the equation:
R = (v^2 * sin(2θ)) / g
Where:
- R is the range of the shell
- v is the initial velocity of the shell
- θ is the angle of elevation
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Given that the angle of elevation (θ) is 30 degrees and the initial velocity (v) is 2000 m/s, we can substitute these values into the equation:
R = (2000^2 * sin(2 * 30)) / 9.8
Let's calculate:
R = (4000000 * sin(60)) / 9.8
R = (4000000 * √3/2) / 9.8
R = (4000000 * 1.732) / 9.8
R ≈ 694897.959 m
Therefore, the range of the shell fired is approximately 694897.959 meters.