A cannon is fired from a cliff 200 m high downward at an angle of 30o with respect to the horizontal. If the muzzle velocity is 34 m/s, what is its speed (in m/s) when it hits the ground?
To find the speed of the cannonball when it hits the ground, we need to analyze its motion in the horizontal and vertical directions separately.
First, let's analyze the motion in the vertical direction. We can use the equations of motion to find the time it takes for the cannonball to hit the ground.
Given:
Initial vertical displacement (s) = -200 m (downward direction)
Vertical acceleration (a) = -9.8 m/s^2 (due to gravity)
Using the equation:
s = ut + (1/2)at^2
where:
s = vertical displacement
u = initial vertical velocity
t = time
Plugging in the values, we have:
-200 = 0t + (1/2)(-9.8)t^2
Simplifying the equation:
-200 = -4.9t^2
Dividing both sides by -4.9:
t^2 = 40.82
Taking the square root of both sides:
t ≈ 6.38 s
Now that we know the time it takes for the cannonball to reach the ground, we can analyze its motion in the horizontal direction.
Since the cannonball is fired at an angle of 30° with respect to the horizontal, we need to find the horizontal component of the velocity.
Given:
Initial velocity (V) = 34 m/s
Angle (θ) = 30°
The horizontal component of the velocity (Vx) can be found using the trigonometric function cosine:
Vx = V * cos(θ)
Plugging in the values, we have:
Vx = 34 * cos(30°)
Vx ≈ 29.47 m/s
Now, we can find the distance traveled (dx) in the horizontal direction using the equation:
dx = Vx * t
Plugging in the values:
dx = 29.47 * 6.38
dx ≈ 187.94 m
Therefore, the speed of the cannonball when it hits the ground is equal to the magnitude of the velocity vector, which is the resultant of the horizontal and vertical components. We can find it using the Pythagorean theorem:
Speed = sqrt(dx^2 + (dy)^2)
where:
dx = horizontal distance = 187.94 m
dy = vertical distance = 200 m (negative because it is downward)
Plugging in the values and calculating:
Speed = sqrt(187.94^2 + (-200)^2)
Speed ≈ sqrt(35280.0436 + 40000)
Speed ≈ sqrt(75280.0436)
Speed ≈ 274.45 m/s
Therefore, the speed of the cannonball when it hits the ground is approximately 274.45 m/s.