A cannon is mounted on a truck that moves forward at a speed of 5 m/s. The operator wants to launch a ball from a cannon so the ball goes as far as possible before hitting the level surface. The muzzle velocity of the cannon is 50 m/s. What angle from the horizontal should the operator point the cannon? (Answer: 49 degrees) Hint: The cannon isn't stationary.
To find the angle at which the cannon should be pointed for the ball to go as far as possible before hitting the level surface, we need to analyze the motion of the ball.
Let's break down the vertical and horizontal components of the ball's motion.
1. Vertical Motion:
The only force acting on the ball vertically is the force of gravity, which pulls the ball downward. The vertical motion can be described using the equations of motion.
The vertical velocity (Vy) of the ball at any time can be calculated using the equation:
Vy = Vy0 - g * t
Where Vy0 is the initial vertical velocity (0 m/s) and g is the acceleration due to gravity (9.8 m/s^2).
Since the ball will hit the level surface at the same vertical height from where it was launched, we can take the time it takes for the ball to hit the ground as the total flight time (t).
2. Horizontal Motion:
The horizontal motion of the ball is influenced by the initial velocity of the ball (Vx) and the velocity of the truck (Vt). Since the truck is moving forward at 5 m/s, the initial horizontal velocity (Vx0) of the ball will be equal to the velocity of the truck.
Vx0 = Vt = 5 m/s
The horizontal displacement of the ball (S) can be calculated using the equation:
S = Vx * t
To maximize the horizontal distance traveled by the ball, we need to find the value of t that gives the maximum horizontal displacement.
Now, recall that the initial velocity of the ball (V0) can be calculated using the initial horizontal and vertical velocities:
V0 = sqrt(Vx0^2 + Vy0^2)
To find the angle (θ) at which the cannon should be pointed, we need to calculate the angle that gives us the desired muzzle velocity (V0).
We can use the equation:
Vx0 = V0 * cos(θ)
Vy0 = V0 * sin(θ)
Substituting Vx0 and Vy0 with their respective equations, we get:
Vt = V0 * cos(θ) (equation 1)
0 = V0 * sin(θ) - g * t (equation 2)
Rearrange equation 1 to solve for V0:
V0 = Vt / cos(θ)
Substitute V0 into equation 2:
0 = (Vt / cos(θ)) * sin(θ) - g * t
Rearrange the equation to isolate t:
t = (Vt / (g * sin(2θ)))
To maximize the horizontal distance, we need to find the value of θ that gives us the longest flight time, t. This can be achieved by setting d(t)/dt = 0, where d(t)/dt is the derivative of t with respect to θ.
Differentiating t with respect to θ, we get:
0 = (Vt / (g * sin(2θ))) * (cos(2θ))
0 = Vt * cos(2θ) / (g * sin(2θ))
Simplify further:
tan(2θ) = 2 * Vt / g
Solve for 2θ:
2θ = arctan(2 * Vt / g)
Finally, divide the obtained value of 2θ by 2 to find θ:
θ = arctan((2 * Vt / g) / 2)
Now, substitute the given values: Vt = 5 m/s and g = 9.8 m/s^2 into the above equation to find the value of θ.