A cannon fires a shell at an angle of 20º to the horizontal. The shell reaches a range of 436 meters, landing at the same elevation as the starting point. What was the initial velocity of the shell
Range = (Vo^2/g)sin40 = 436 m
Solve for Vo
Yes, that is sin 40, not sin 20 in the equation. That is because of the trigonometric identity
2 sinx cosx = sin(2x)
Derive the range and you will see why both sinx and cosx appear.
To find the initial velocity of the shell, we can break down the motion into horizontal and vertical components.
First, let's consider the horizontal component. The horizontal motion of the shell is not affected by gravity, so the initial horizontal velocity will remain constant throughout its flight.
The horizontal velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
Where:
Vx is the horizontal velocity
V is the initial velocity of the shell
θ is the launch angle
In this case, θ is 20º. So, we have:
Vx = V * cos(20º)
Now, let's consider the vertical component. The vertical motion of the shell is affected by gravity. The range of the shell is the horizontal distance it travels before landing at the same elevation as the starting point. The range (R) can be found using the equation:
R = (V^2 * sin(2θ)) / g
Where:
R is the range
V is the initial velocity of the shell
θ is the launch angle
g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, θ is 20º and R is 436 meters. So, we have:
436 = (V^2 * sin(2 * 20º)) / 9.8
Now, we have two equations:
Vx = V * cos(20º)
436 = (V^2 * sin(2 * 20º)) / 9.8
We can solve these two equations simultaneously to find the initial velocity of the shell (V).