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
To find the angle θ, we can use the fact that the horizontal component of the velocity remains constant throughout the motion. Let's break down the problem into steps:
Step 1: Find the time it takes for the ball to hit the ground.
We can use the equation of motion for vertical motion:
h = vit + (1/2)gt^2
Where:
h = height (50 m)
vi = initial vertical velocity (0 m/s, since the ball is fired horizontally)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the values, we have:
50 = 0t + (1/2)(-9.8)t^2
50 = -4.9t^2
Rearranging the equation, we get:
4.9t^2 = 50
t^2 = 50 / 4.9
t ≈ 3.19 seconds
Step 2: Find the horizontal distance traveled by the ball.
Since the horizontal component of the velocity remains constant, we can use the equation:
d = vht
Where:
d = horizontal distance
vh = horizontal velocity (75 m/s)
t = time (3.19 seconds)
Plugging in the values, we have:
d = (75 m/s)(3.19 s)
d ≈ 239.25 meters
Step 3: Find the final vertical velocity of the ball.
Using the same equation of motion for vertical motion:
v = vi + gt
Where:
v = final vertical velocity
vi = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (3.19 seconds)
Plugging in the values, we have:
v = (0 m/s) + (-9.8 m/s^2)(3.19 s)
v ≈ -31.262 m/s
Step 4: Find the resultant velocity of the ball.
The resultant velocity is the vector sum of the horizontal and vertical velocities. Using the Pythagorean theorem:
v^2 = (vh)^2 + (vv)^2
Where:
v = resultant velocity
vh = horizontal velocity (75 m/s)
vv = vertical velocity (-31.262 m/s)
Plugging in the values, we have:
v^2 = (75 m/s)^2 + (-31.262 m/s)^2
v ≈ 81.9 m/s
Step 5: Find the angle θ with the horizontal.
Using trigonometry, we can use the inverse tangent function:
θ = tan^(-1)(vv/vh)
Plugging in the values, we have:
θ = tan^(-1)(-31.262 m/s / 75 m/s)
θ ≈ -22.8 degrees
Note: The negative sign is used to indicate that the angle is below the horizontal. Alternatively, we can use the absolute value of the angle, which gives us θ ≈ 22.8 degrees.
Therefore, the angle θ is approximately 22.8 degrees below the horizontal.
To find the angle θ, we can use the principle of conservation of mechanical energy.
When the cannonball is fired horizontally, it has only kinetic energy. At the top of the building, it has potential energy due to its height. When it hits the ground, it has a combination of kinetic and potential energy.
The total mechanical energy of the cannonball is conserved throughout its motion, ignoring air resistance. This means that the initial mechanical energy (kinetic + potential) is equal to the final mechanical energy (kinetic + potential).
At the top of the building, the cannonball only has potential energy, which can be calculated using the formula:
Potential energy = mass x gravitational acceleration x height
Since the cannonball is fired horizontally, its vertical motion is not affected by the initial velocity. Therefore, at the top of the building, the cannonball only has potential energy.
When the cannonball hits the ground, it has both kinetic and potential energy. The potential energy is now zero since it is at ground level, and the kinetic energy can be calculated using the formula:
Kinetic energy = (1/2) x mass x velocity^2
By equating the initial mechanical energy to the final mechanical energy, we can solve for the final velocity of the cannonball at the ground level.
Since the angle θ is the angle between the horizontal and the velocity vector of the cannonball at the ground level, we can use trigonometry to find θ.
The horizontal velocity component remains constant throughout the motion, so we can use the equation for horizontal motion to find the time it takes for the cannonball to hit the ground when fired horizontally:
Distance = velocity (horizontal) x time
In this case, the distance is the horizontal distance the cannonball travels from the top of the building to the ground, which is the same as the horizontal distance from where it was fired, since there is no horizontal acceleration.
Once we have the time it takes for the cannonball to hit the ground, we can use the vertical motion equation to find the vertical velocity component at the time of impact:
Vertical velocity = gravitational acceleration x time
Now that we have both the horizontal and vertical components of the velocity at the time of impact, we can use trigonometry to find θ.
The tangent of θ is given by the ratio of the vertical velocity component to the horizontal velocity component.
θ = arctan(vertical velocity / horizontal velocity)