The range of a target is found to be 20km.A shell leaves again with a velocity of 500ms-1.what must be the angle of elevation of the gun.if the ground is leveled.
To find the angle of elevation of the gun, we can use the range formula for projectile motion. The formula is:
Range = (v^2 * sin(2θ)) / g
Where:
- Range is the horizontal distance covered by the projectile (in this case, 20 km or 20,000 m).
- v is the initial velocity (in this case, 500 m/s).
- θ is the angle of elevation.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the formula, we can solve for θ:
sin(2θ) = (Range * g) / v^2
Now, let's substitute the given values into the equation:
sin(2θ) = (20,000 * 9.8) / (500^2)
sin(2θ) = 196,000 / 250,000
sin(2θ) = 0.784
To find the angle θ, we need to take the inverse sine (or arcsine) of 0.784:
θ = arcsin(0.784)
Using a calculator, we find:
θ ≈ 51.11 degrees
Therefore, the angle of elevation of the gun should be approximately 51.11 degrees.
To find the angle of elevation of the gun, we need to use the concept of projectile motion. The range of a projectile can be determined using the formula:
Range = (v^2 * sin(2θ))/g
Where:
- Range is the horizontal distance covered by the projectile
- v is the initial velocity of the projectile
- θ is the angle of elevation
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the given range is 20 km, which we need to convert to meters:
20 km = 20,000 m
The given initial velocity is 500 m/s, acceleration due to gravity is 9.8 m/s^2, and we need to find the angle of elevation (θ).
Let's rearrange the formula to solve for θ:
Range = (v^2 * sin(2θ))/g
=> 20,000 = (500^2 * sin(2θ))/9.8
=> 20,000 = (250,000 * sin(2θ))/9.8
Now, we can solve for sin(2θ):
sin(2θ) = (20,000 * 9.8) / 250,000
=> sin(2θ) = 0.784
To find 2θ, we can take the inverse sine (sin^(-1)) of 0.784:
2θ = sin^(-1)(0.784)
θ = (sin^(-1)(0.784))/2
Using a calculator, we find that sin^(-1)(0.784) is approximately 50.88 degrees.
Hence, the angle of elevation of the gun should be approximately 50.88 degrees.
20km=timeinair*500*cosAngle
solve for time in air.
in the vertical
hf=hi+vi*timeinair-1/2 *(9.8)timeinair^2
or time in air=500*sinAngle/4.9
solve now for the angle.
Range = Vo^2*sin(2A)/g = 20,000 m.
500^2*sin(2A)/9.8 = 20.000,
2A =
A =