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?
Ignoring air-resistance, equate energies:
(PE+KE)initial = (PE+KE)final
mgh+(1/2)mu² = (1/2)mv²
Solve for v.
I get approx. 71 m/s.
thank you so much!
You're welcome!
To find the speed of the cannonball when it hits the ground, we first need to break down its initial velocity into horizontal and vertical components.
Given:
- Initial height of the cliff, h = 200 m
- Launch angle with respect to the horizontal, θ = 30°
- Muzzle velocity, v₀ = 34 m/s
1. Calculate the initial vertical velocity (v₀y)
v₀y = v₀ * sin(θ)
v₀y = 34 m/s * sin(30°)
v₀y ≈ 17 m/s
2. Calculate the time it takes for the cannonball to reach the ground.
We can use the equation for vertical motion to find the time of flight (t).
h = v₀y * t - (1/2) * g * t² (where g = acceleration due to gravity ≈ 9.8 m/s²)
200 m = 17 m/s * t - (1/2) * 9.8 m/s² * t²
Rearranging the equation to solve for t:
(1/2) * 9.8 m/s² * t² - 17 m/s * t + 200 m = 0
Solving this quadratic equation using the quadratic formula will give us the value of t.
3. Once we have the time of flight (t), we can calculate the horizontal distance traveled (x) using the equation:
x = v₀x * t (where v₀x is the initial horizontal velocity)
Since the cannonball is launched horizontally, the horizontal velocity is equal to the muzzle velocity:
v₀x = v₀ = 34 m/s
4. Calculate the speed of the cannonball when it hits the ground.
Now that we have the horizontal distance (x) and time of flight (t), we can calculate the speed at impact using the formula for average speed:
Speed = Total distance / Time = x / t
By following these steps and performing the calculations, we can determine the speed (in m/s) of the cannonball when it hits the ground.