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 can use the horizontal and vertical components of its initial velocity.
1. First, let's find the initial vertical velocity component (Vy0):
Vy0 = V0 * sin(theta)
where V0 is the muzzle velocity and theta is the angle with respect to the horizontal.
Vy0 = 34 m/s * sin(30°)
Vy0 ≈ 17 m/s
2. Next, let's find the time it takes for the cannonball to reach the ground. We can use the equation:
h = Vy0 * t + (1/2) * g * t^2
where h is the height of the cliff, t is the time, and g is the acceleration due to gravity.
Rearranging the equation, we have:
200 m = 17 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation, we get a quadratic equation:
4.9 t^2 + 17 t - 200 = 0
We can solve this equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
a = 4.9, b = 17, c = -200
Plugging the values into the formula:
t = (-17 ± sqrt(17^2 - 4*4.9*(-200))) / 2*4.9
t ≈ 4.36 s (ignoring the negative solution)
3. Finally, we can find the horizontal component of the cannonball's speed (Vx) using the equation:
Vx = V0 * cos(theta)
where V0 is the muzzle velocity and theta is the angle with respect to the horizontal.
Vx = 34 m/s * cos(30°)
Vx ≈ 29.5 m/s
4. Now that we know the horizontal velocity (Vx) and the time (t) it takes for the cannonball to hit the ground, we can find the horizontal distance traveled (X) using the equation:
X = Vx * t
X = 29.5 m/s * 4.36 s
X ≈ 128.22 m
5. The speed of the cannonball when it hits the ground is the magnitude of its velocity, which can be found using the Pythagorean theorem:
v = sqrt(Vx^2 + Vy^2)
v = sqrt((29.5 m/s)^2 + (17 m/s)^2)
v ≈ 34.10 m/s
Therefore, the speed of the cannonball when it hits the ground is approximately 34.10 m/s.
To find the speed of the cannon when it hits the ground, we need to analyze the horizontal and vertical components of its motion separately.
First, let's calculate the time it takes for the cannonball to hit the ground.
We know that the initial vertical velocity is given by v₀y = v₀ * sin(theta), where v₀ is the initial velocity and theta is the angle of projection.
v₀y = 34 m/s * sin(30°)
v₀y = 34 m/s * 0.5
v₀y = 17 m/s
Next, we can use the kinematic equation for vertical motion:
y = v₀yt + (1/2)gt²
Where:
y is the vertical displacement (200 m downwards, as the cliff is 200 m high)
v₀y is the initial vertical velocity (17 m/s)
g is the acceleration due to gravity (approximately 9.8 m/s²)
t is the time taken
Rearranging the equation, we have:
200 m = 17 m/s * t + (1/2) * 9.8 m/s² * t²
200 m = 17t + 4.9t²
This is a quadratic equation, so we can solve it by setting it equal to zero:
4.9t² + 17t - 200 = 0
Using the quadratic formula, t = (-b ± sqrt(b² - 4ac)) / (2a),
where a = 4.9, b = 17, and c = -200.
Solving this equation, we get two possible values for t: t1 = 4.36 s and t2 = -10.16 s.
Since time cannot be negative, the cannon will hit the ground after 4.36 seconds.
Now, let's calculate the horizontal distance traveled by the cannon:
We know that the horizontal velocity remains constant throughout the motion. Hence, the horizontal distance traveled (d) is given by:
d = v₀x * t
where v₀x is the horizontal component of the initial velocity, which can be calculated using v₀x = v₀ * cos(theta).
v₀x = 34 m/s * cos(30°)
v₀x = 34 m/s * 0.866
v₀x = 29.524 m/s
Now we can substitute the values into the equation:
d = 29.524 m/s * 4.36 s
d ≈ 128.6 m
The horizontal distance traveled by the cannonball is approximately 128.6 meters.
Finally, we can find the speed at which the cannon hits the ground by calculating the resultant velocity, which is the vector sum of the horizontal and vertical components of motion. We can use the Pythagorean theorem:
v = sqrt(vx² + vy²)
where vx is the horizontal velocity (v₀x) and vy is the vertical velocity (v₀y).
v = sqrt((29.524 m/s)² + (17 m/s)²)
v ≈ 34.06 m/s
Therefore, the speed of the cannon when it hits the ground is approximately 34.06 m/s.