A cannon ball is fired at 50 m/s at a 37 degree angle above the horizontal. Calculate the time the cannon ball in the air, calculate how high it rises and calculate how far from the cannon the ball lands?
Answers: 6.14 s, 46.2 m, 245.2 m
To calculate the time the cannonball is in the air, you can use the vertical motion equation that relates the time of flight, initial vertical velocity, and the acceleration due to gravity.
1. First, split the initial velocity into its horizontal and vertical components. The horizontal component is given by \(V_{x} = V \cdot cos(\theta)\), where \(V\) is the initial velocity and \(\theta\) is the angle above the horizontal.
In this case, \(V = 50 m/s\) and \(\theta = 37\) degrees.
So, \(V_{x} = 50 \cdot cos(37)\).
2. Next, calculate the time it takes for the object to reach the maximum height using the formula \(t_{\text{rise}} = \frac{V_{y}}{g}\), where \(V_{y}\) is the vertical component of the initial velocity (or the initial vertical velocity) and \(g\) is the acceleration due to gravity, which is approximately \(9.8 \, \text{m/s}^2\).
To find the vertical component of the initial velocity, use \(V_{y} = V \cdot sin(\theta)\) and substitute the given values: \(V_{y} = 50 \cdot sin(37)\).
So, \(t_{\text{rise}} = \frac{V_{y}}{g}\).
3. After reaching the maximum height, the projectile will descend back down. The total time of flight, \(t_{\text{total}}\), is twice the time it took to reach the maximum height, so \(t_{\text{total}} = 2 \cdot t_{\text{rise}}\).
Now you can calculate the time the cannonball is in the air by substituting the values into the above equations.
To calculate how high the cannonball rises, you can use the equation \(h = V_{y} \cdot t_{\text{rise}} - \frac{1}{2} \cdot g \cdot t_{\text{rise}}^2\).
To calculate how far from the cannon the ball lands, you can use the equation \(d = V \cdot cos(\theta) \cdot t_{\text{total}}\).
Substituting the values into these equations will give you the answers:
\(t_{\text{total}} = 2 \cdot t_{\text{rise}}\)
\(h = V_{y} \cdot t_{\text{rise}} - \frac{1}{2} \cdot g \cdot t_{\text{rise}}^2\)
\(d = V \cdot cos(\theta) \cdot t_{\text{total}}\)
After substituting the given values, you should obtain the answers: 6.14 s for the time of flight, 46.2 m for the height, and 245.2 m for the distance from the cannon.
G:
Vi = 50m/s
Degree = 37
U: Vy, Vx, Dy, Dx, T
E:
Vy = Vi(Sin(θ)),
Vx = Vi(Cos(θ),
Dy = Vy(1/2)(t) + 1/2(a)(1/2t^2)
Dx = Vx(t) (Because the speed is constant)
t = 6.14 s, h = 46.2, Δx = 245.2 m