Am i doing this right?
A cannon is shot at a 30 degree angle above the horizontal with a velocity of 240 m/s. What is the range of this shot?
Range = Vm^2 sin2(theta) / (g)
(240m/s^2)sin(2(30)) / (10ms^2) = 4988 meters?
Looks good to me.
except that v^2 = (240m/s)^2, not 240m/s^2, and g=10m/s^2, not ms^2
watch out fer them unit thingees.
To calculate the range of the shot, you are correct in using the formula:
Range = (V^2 * sin(2θ)) / g
Where:
- V is the initial velocity of the cannonball (240 m/s in this case)
- θ is the angle of launch (30 degrees in this case)
- g is the acceleration due to gravity (approximately 10 m/s^2)
First, let's calculate sin(2θ):
sin(2θ) = sin(2 * 30) = sin(60) = √3/2
Now substitute the values into the formula:
Range = (240^2 * (√3/2)) / 10
Evaluating the equation:
Range = (57600 * √3/2) / 10
= (57600√3) / 20
= 2880√3
So, the range of the shot would be approximately 4988 meters.
Therefore, your calculation is indeed correct.