. 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 θ with the
horizontal. What is θ ?
how long does it take to fall 50m?
4.9t^2 = 50
t = 3.19 sec
v = at = 9.8*3.19 = 31.3 m/s
So, when it hits, it has
Vx = 75
Vy = 31.3
tanθ = 31.3/75 = 0.4173
θ = 22.6°
To find the angle θ that the velocity vector makes with the horizontal when the ball hits the ground, we can use the concept of projectile motion.
Step 1: Determine the time it takes for the cannonball to hit the ground.
The vertical distance traveled by the cannonball is the height of the building, which is 50 m. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Using the equation s = ut + (1/2)at^2, where s is the vertical distance, u is the initial vertical velocity, a is the acceleration, and t is the time, we can solve for t:
50 = 0t + (1/2)(9.8)t^2
50 = 4.9t^2
t^2 = 50/4.9
t^2 ≈ 10.204
t ≈ √10.204
t ≈ 3.19 s (rounded to two decimal places)
Step 2: Calculate the horizontal distance traveled by the cannonball.
The horizontal distance traveled by the cannonball can be calculated using the equation d = vt, where d is the horizontal distance, v is the initial horizontal velocity, and t is the time.
Since the cannonball is fired horizontally, the initial horizontal velocity is equal to the initial speed, which is 75 m/s.
d = (75 m/s)(3.19 s)
d ≈ 239.25 m (rounded to two decimal places)
Step 3: Calculate the magnitude of the velocity when the cannonball hits the ground.
The magnitude of the velocity vector can be found using the equation v = √((v_x)^2 + (v_y)^2), where v_x is the horizontal velocity and v_y is the vertical velocity.
The vertical velocity can be calculated using the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration, and t is the time.
Since the ball is free-falling vertically, the initial vertical velocity is 0 m/s.
v = 0 + (9.8 m/s^2)(3.19 s)
v ≈ 31.262 m/s (rounded to three decimal places)
The horizontal velocity remains constant throughout the motion, so the horizontal velocity is 75 m/s.
Therefore, the magnitude of the velocity when the cannonball hits the ground is √((75)^2 + (31.262)^2) ≈ 81.871 m/s (rounded to three decimal places).
Step 4: Find the angle θ using trigonometric functions.
θ can be found using the equation tan(θ) = (v_y)/(v_x), where θ is the angle and v_y and v_x are the vertical and horizontal components of the velocity, respectively.
θ = tan^(-1)((31.262)/(75))
θ ≈ 23.57° (rounded to two decimal places)
Therefore, the angle θ that the velocity vector makes with the horizontal when the ball hits the ground is approximately 23.57°.
To find the angle θ that the velocity vector makes with the horizontal when the cannonball hits the ground, we can use the concept of projectile motion.
Let's break down the problem into different components:
1. Vertical motion: The cannonball is dropped vertically from a height of 50 m, which means it initially has a vertical velocity of 0 m/s. The only force acting on it is gravity, which causes it to accelerate downwards at a rate of 9.8 m/s^2.
Using the vertical motion equation:
y = y0 + v0y * t + (1/2) * a * t^2
where:
y = final vertical position (0 m since it hits the ground)
y0 = initial vertical position (50 m)
v0y = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Simplifying the equation, we get:
50 = 0 * t + (1/2) * (-9.8) * t^2
50 = -4.9 * t^2
Solving for t, we find that t ≈ 3.18 s.
2. Horizontal motion: The cannonball is fired horizontally with an initial speed of 75 m/s. Since there is no vertical acceleration in the horizontal direction, the cannonball will continue to move at a constant horizontal velocity throughout its flight.
The horizontal distance traveled is given by:
x = v0x * t
where:
x = horizontal distance traveled (unknown)
v0x = initial horizontal velocity (75 m/s)
t = time of flight (3.18 s)
Substituting the values, we find:
x = 75 * 3.18
x ≈ 238.5 m
3. Finding θ: Now that we know the vertical and horizontal distances traveled, we can find the angle θ using the tangent function.
tan(θ) = vertical distance / horizontal distance
tan(θ) = 50 / 238.5
Using inverse tangent (tan^-1 or arctan), we find:
θ ≈ 12.3 degrees
Therefore, the velocity vector makes an angle θ ≈ 12.3 degrees with the horizontal when the cannonball hits the ground.