A cannon is fired at 30 degrees above the horizontal with a velocity of 200 m/s from the edge of a 125 m high cliff. Calculate where the cannonball lands on the level plain below.
just plug your numbers into
y = tanθ x - g/(2(v cosθ)^2) x^2
and solve for x when y = -125
Well, I hope that cannonball has a good sense of direction because it's going to take quite the plunge! Let's crunch some numbers and see where it ends up.
First, we need to break down the initial velocity of the cannonball into its horizontal and vertical components. Since the cannonball is fired at a 30-degree angle above the horizontal, we can find the horizontal component by multiplying the initial velocity (200 m/s) by the cosine of the angle (30 degrees).
Horizontal component = 200 m/s * cos(30 degrees) = 173.21 m/s
Now, let's find out how long it takes for the cannonball to hit the ground. We'll focus on the vertical motion here. We can use the formula:
y = yo + vo*t - (1/2)*g*t^2
where:
y = vertical displacement (0 m, since it hits the ground)
yo = initial vertical position (125 m)
vo = initial vertical velocity (computed using the initial velocity and sine of the angle)
g = acceleration due to gravity (9.8 m/s^2)
t = time it takes to hit the ground (what we're trying to find)
Plugging in our values, the equation becomes:
0 = 125 m + vo*t - (1/2)*9.8 m/s^2*t^2
Simplifying further, we have:
(1/2)*9.8 m/s^2*t^2 - vo*t - 125 m = 0
Since this is a quadratic equation, we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4*a*c))/(2*a)
where:
a = (1/2)*9.8 m/s^2 = 4.9 m/s^2
b = -vo = -200 m/s * sin(30 degrees) = -100 m/s
c = -125 m
Plugging in our values, we find:
t = (-(-100) ± √((-100)^2 - 4*(4.9)*(-125)))/(2*(4.9))
After all the math, the two solutions for time will be:
t = 11.96 s (approx.) or t = -2.09 s (approx.)
Since time can't be negative, we discard the negative solution.
Now, let's find the horizontal distance the cannonball travels using the horizontal component and the time:
Horizontal distance = horizontal component * time = 173.21 m/s * 11.96 s = 2072.49 m (approx.)
Therefore, the cannonball lands approximately 2072.49 meters away from the edge of the cliff. Just be sure to stand clear when it comes crashing down!
To calculate where the cannonball lands on the level plain below, we need to determine the horizontal and vertical components of its initial velocity.
1. Horizontal Component:
The horizontal component of the initial velocity remains constant throughout the projectile's motion. We can find it using trigonometry:
Horizontal component = Velocity * cos(angle)
Given:
Velocity = 200 m/s
Angle = 30 degrees
Horizontal component = 200 * cos(30)
Horizontal component ≈ 173.2 m/s
2. Vertical Component:
The vertical component of the initial velocity changes due to the effect of gravity. We can find it using trigonometry:
Vertical component = Velocity * sin(angle)
Vertical component = 200 * sin(30)
Vertical component ≈ 100 m/s
Now, we need to calculate the time it takes for the cannonball to hit the ground.
3. Time of Flight:
The time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity (g ≈ 9.8 m/s²):
Time of flight = (2 * Vertical component) / g
Time of flight ≈ (2 * 100) / 9.8
Time of flight ≈ 20.41 s
The horizontal distance the cannonball travels can be calculated using the horizontal component of the initial velocity and the time of flight:
4. Horizontal Distance:
Horizontal distance = Horizontal component * Time of flight
Horizontal distance ≈ 173.2 * 20.41
Horizontal distance ≈ 3,540 m
Therefore, the cannonball lands approximately 3,540 meters away from the edge of the cliff on the level plain below.
To calculate where the cannonball lands on the level plain below, we need to break down the problem into two components: horizontal and vertical.
First, let's calculate the time it takes for the cannonball to reach the ground. Since we are ignoring air resistance, the vertical motion of the cannonball can be analyzed independently.
Using the equation of motion for vertical displacement:
d = vit + (1/2)gt^2
where:
d = vertical displacement (125 m, as the cannon is fired from the edge of a 125 m high cliff)
vi = initial vertical velocity (0 m/s, as the cannonball is not moving vertically at the start)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Simplifying the equation, we get:
125 = (1/2)(-9.8)t^2
Solving for t, we get:
t^2 = (125 * 2) / 9.8
t^2 = 256.12
t ≈ 16 seconds (taking the square root)
Now, let's determine the horizontal distance traveled by the cannonball during this time. We can use the equation of motion for horizontal displacement:
d = vix * t
where:
d = horizontal displacement (the impact point we are trying to determine)
vix = initial horizontal velocity (v * cosθ, where v is 200 m/s and θ is the launch angle of 30 degrees)
t = time (16 seconds)
Plugging in the values, we get:
d = (200 * cos30°) * 16
Evaluating this expression, we find:
d ≈ 2773.86 meters
Therefore, the cannonball lands approximately 2773.86 meters away from the cliff, on the level plain below.