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 resolve the initial velocity into its horizontal and vertical components.
Given:
Muzzle velocity (V₀) = 34 m/s
Angle of projection (θ) = 30°
Height of the cliff (h) = 200 m
First, let's find the vertical component of the velocity (Vy₀):
Vy₀ = V₀ * sin(θ)
= 34 m/s * sin(30°)
= 17 m/s
Next, we can find the time taken for the cannonball to hit the ground by using the following equation for vertical motion:
h = (1/2) * g * t²
where:
h = height of the cliff = 200 m
g = acceleration due to gravity = 9.8 m/s²
t = time taken to hit the ground
Rearranging the equation:
t² = 2h / g
t = √(2h / g)
t = √(2 * 200 m / 9.8 m/s²)
t ≈ √40.82
t ≈ 6.39 s
Now, we can find the horizontal component of the velocity (Vx₀) using the following equation:
Vx₀ = V₀ * cos(θ)
= 34 m/s * cos(30°)
= 29.47 m/s
Since there is no horizontal acceleration, the horizontal component of the velocity remains constant throughout the motion.
Finally, we can find the speed (Vf) of the cannonball when it hits the ground:
Vf = √(Vx₀² + Vy₀²)
= √((29.47 m/s)² + (17 m/s)²)
= √(867.60 + 289)
= √1156.60
≈ 34.04 m/s
Therefore, the speed of the cannonball when it hits the ground is approximately 34.04 m/s.