A cannon ball is fired horizontally with an initial speed of 75 m/s from the top of a 50 m tall building. When the ball hits the ground, its velocity vector makes an angle with the horizontal. What is the angle?
Xo = 75 m/s.
Y^2 = Yo^2 + 2g*h
Y^2 = 0 + 19.6*50 = 980
Y = 31.3 m/s.
V = 75 + 31.3i
tan A = Y/Xo = 31.3/75 = 0.41740
A = 22.7o
To find the angle made by the velocity vector of the cannonball with the horizontal when it hits the ground, we can use the principles of projectile motion.
Let's break down the problem step by step:
1. First, we need to find the time it takes for the cannonball to hit the ground. Since the initial velocity of the cannonball is horizontal, it will not change throughout its motion. The only force acting on it is gravity, causing it to accelerate downward. Therefore, we can use the equation for vertical displacement in free fall to find the time it takes to fall from a height of 50 m.
The equation for vertical displacement is: Δy = v₀y × t + 0.5 × a × t²
Since the cannonball is fired horizontally, its initial vertical velocity (v₀y) is 0 m/s, as all its initial velocity is directed horizontally. The acceleration due to gravity (a) is approximately 9.8 m/s².
Plugging in the values, we have:
50 m = 0 × t + (0.5 × 9.8 × t²)
Rearranging the equation, we get:
0.5 × 9.8 × t² = 50
Simplifying, we get:
4.9 × t² = 50
Dividing both sides by 4.9, we get:
t² = 10
Taking the square root of both sides, we find:
t = √10 ≈ 3.16 s
So, it takes approximately 3.16 seconds for the cannonball to hit the ground.
2. Next, we need to find the horizontal distance covered by the cannonball during this time. Since the initial velocity of the cannonball is 75 m/s and it travels for 3.16 seconds horizontally, we can use the equation for horizontal displacement to find the distance.
The equation for horizontal displacement is: Δx = v₀x × t
Since the cannonball is fired horizontally, its initial horizontal velocity (v₀x) is 75 m/s. Plugging in the values, we have:
Δx = 75 m/s × 3.16 s
Calculating the value, we find:
Δx ≈ 237.0 m
Therefore, the cannonball covers a horizontal distance of approximately 237.0 meters.
3. Now that we have the vertical and horizontal distances covered by the cannonball, we can find the tangent of the angle between the velocity vector and the horizontal axis using the equation: tan(θ) = Δy / Δx.
Plugging in the values, we have:
tan(θ) = 50 m / 237.0 m
Calculating the value, we find:
tan(θ) ≈ 0.211
4. Finally, to find the angle (θ), we can take the inverse tangent (arctan) of both sides of the equation.
Using a calculator, we find:
θ ≈ arctan(0.211)
Calculating the value, we find:
θ ≈ 12.0°
So, the angle made by the velocity vector of the cannonball with the horizontal when it hits the ground is approximately 12.0 degrees.