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?
A ball is dropped from rest from 10 m above ground. The speed of the ball in the middle of trip to ground is close to
Answer
14 m/s
10 m/s
can't be determined because mass is not given
100 m/s
To determine the angle at which the operator should point the cannon, we need to find the optimal launch angle that maximizes the range of the ball. The range is the horizontal distance covered by the ball before hitting the ground.
The range of a projectile can be calculated using the following formula:
Range = (Velocity^2 * sin(2*θ)) / g
Where:
- Range is the horizontal distance covered by the projectile.
- Velocity is the initial velocity, i.e., the muzzle velocity of the cannon.
- θ is the launch angle.
- g is the acceleration due to gravity, which is approximately 9.8 m/s² near the surface of the Earth.
Since the truck is moving, we need to consider the relative velocity of the ball with respect to the ground. The relative velocity is the vector sum of the truck's velocity and the ball's velocity.
In this case, the relative velocity of the ball with respect to the ground is the sum of the truck's velocity and the ball's muzzle velocity. So, the total initial velocity of the ball can be obtained by adding the truck's velocity and the muzzle velocity of the cannon:
Total Velocity = Truck Velocity + Muzzle Velocity
Now, we can substitute this total velocity into the formula for the range:
Range = (Total Velocity^2 * sin(2*θ)) / g
To find the launch angle that maximizes the range, we can differentiate the range formula with respect to θ and set it to zero:
d(Range)/d(θ) = ((Total Velocity^2 * 2 * cos(2*θ)) / g) = 0
Simplifying, we get:
cos(2*θ) = 0
Using the double angle formula for cosine, we have:
2*cos^2(θ) - 1 = 0
Hence, cos^2(θ) = 1/2
Taking the square root of both sides gives:
cos(θ) = ± √(1/2)
Solving for θ, we find:
θ = ± 45 degrees
Since the launch angle cannot be negative, the only valid solution is:
θ = 45 degrees
Therefore, the operator should point the cannon at an angle of 45 degrees from the horizontal to launch the ball as far as possible before hitting the level surface.