A projectile is fired from a cliff 300 m high with an initial velocity of 400 m/s. If the firing angle is 30 degrees from the horizontal, compute the horizontal range of the projectile.

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Well, let's see. A projectile is fired from a cliff... why did it leave? Did it have some kind of mid-life crisis? Anyway, it was probably looking for a higher ground. So, it starts at a cliff that's 300 m high and takes off at an angle of 30 degrees from the horizontal. The initial velocity is 400 m/s.

Now, to find the horizontal range, we need to first split the initial velocity into its horizontal and vertical components. The horizontal component can be found using some trigonometry. In this case, the clown trigonometry.

So, if the angle is 30 degrees, that makes it half of 60 degrees. And half of 60 is... 30! Hey, it's the same angle. So, we have a nice pair of matching angles here. Isn't that cute?

Since we have a 30-degree angle, we can use the clown trigonometric formula:

cos(30) = adjacent/hypotenuse

The adjacent side is the horizontal component of the velocity, and the hypotenuse is the magnitude of the velocity, which is 400 m/s. So, just plug in the values and solve for the adjacent side. Go on, I'll wait.

Got it? Great! The horizontal component of the velocity turns out to be approximately 346.41 m/s.

Now, to find the time it takes for the projectile to hit the ground, we need to look at the vertical motion. But hey, don't worry, we don't want to get too high.

We can use another clown trigonometric formula to find the time of flight. Do you know which one? It's:

sin(theta) = opposite/hypotenuse

In this case, the opposite side is the vertical component of the velocity, which turns out to be 200 m/s (sin(30) x 400 m/s).

Now, divide the initial vertical velocity by the gravitational acceleration (which we'll assume is 9.8 m/s^2). That will give you the time it takes for the projectile to reach the highest point.

Twinkle, twinkle, little projectile, now it's descending to the ground. But how long does it take? To find the total time of flight, we double the time it took to reach the highest point. Why double? Well, the projectile needs to come back down, right?

Now that we know the time of flight, we can find the horizontal range by multiplying the horizontal component of the velocity by the time of flight.

So, let me calculate this for you... a little drum roll please...

The horizontal range of the projectile is approximately 346.41 m/s multiplied by the total time of flight.

To compute the horizontal range of the projectile, we need to find the horizontal distance covered by the projectile before it hits the ground.

First, let's break down the initial velocity into its horizontal and vertical components:

Horizontal component: v₀x = v₀ * cos(θ)
Vertical component: v₀y = v₀ * sin(θ)

where
v₀ = initial velocity = 400 m/s
θ = firing angle = 30 degrees

Substituting the given values, we can find the horizontal and vertical components of the initial velocity:

v₀x = 400 m/s * cos(30°)
= 400 m/s * √3/2
≈ 346.41 m/s

v₀y = 400 m/s * sin(30°)
= 400 m/s * 1/2
= 200 m/s

Now, we can calculate the time it takes for the projectile to reach the ground. Since the only force acting on the projectile is gravity, the vertical motion can be described using the equation:

y = v₀yt - (1/2)gt²

where
y = vertical displacement = -300 m (negative because the projectile is moving downward)
v₀y = vertical component of initial velocity = 200 m/s
g = acceleration due to gravity = 9.8 m/s²
t = time

Substituting the given values, we can solve for t:

-300 m = (200 m/s)t - (1/2)(9.8 m/s²)t²
-300 m = 200t - 4.9t²

Rearranging the equation to the form 4.9t² - 200t - 300 = 0, we can solve for t using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

where
a = 4.9
b = -200
c = -300

t = (200 ± √((-200)² - 4 * 4.9 * (-300))) / (2 * 4.9)
t ≈ (200 ± 641.04) / 9.8

The positive value of t represents the time when the projectile hits the ground. So:

t ≈ (200 + 641.04) / 9.8 ≈ 84.67 s

Now that we know the time of flight, we can find the horizontal distance covered by the projectile:

Range = v₀x * t

Range = 346.41 m/s * 84.67 s
≈ 29,293.34 m

Therefore, the horizontal range of the projectile is approximately 29,293.34 m.

To find the horizontal range of the projectile, we need to determine the time it takes for the projectile to reach the ground.

First, let's break down the initial velocity into horizontal and vertical components. The horizontal component is given by Vx = V * cos(theta) and the vertical component is given by Vy = V * sin(theta), where V is the initial velocity and theta is the firing angle.

In this case, V = 400 m/s and theta = 30 degrees, so we can calculate:

Vx = 400 m/s * cos(30 degrees) = 400 m/s * 0.866 = 346.41 m/s
Vy = 400 m/s * sin(30 degrees) = 400 m/s * 0.5 = 200 m/s

The initial vertical velocity is 200 m/s, and we need to find the time it takes for the projectile to reach the ground, so we can use the formula: h = Vy * t + (1/2) * g * t^2, where h is the displacement in the vertical direction, Vy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).

In this case, h = -300 m (taking the downward direction as negative), Vy = 200 m/s, and g = -9.8 m/s^2, so we have:

-300 m = 200 m/s * t + (1/2) * (-9.8 m/s^2) * t^2

Simplifying this quadratic equation, we get:
-4.9 t^2 + 200 t - 300 = 0

We can solve this quadratic equation to find the time t. Using the quadratic formula, t = (-b ± sqrt(b^2 - 4ac)) / 2a, where a = -4.9, b = 200, and c = -300.

Substituting these values into the formula, we get:
t = (-200 ± sqrt(200^2 - 4 * -4.9 * -300)) / 2 * -4.9

Calculating this expression, we have:
t ≈ (-200 ± sqrt(40000 - 5880)) / -9.8
t ≈ (-200 ± sqrt(34120)) / -9.8

Now, we have two possible values for t, one positive and one negative. The negative value does not make physical sense in this context, so we take the positive value:

t ≈ (-200 + sqrt(34120)) / -9.8 ≈ 22.13 s

So, it takes approximately 22.13 seconds for the projectile to reach the ground.

Finally, we can find the horizontal range by multiplying the horizontal velocity (Vx) by the time (t):

horizontal range = Vx * t
horizontal range = 346.41 m/s * 22.13 s
horizontal range ≈ 7651.75 m or 7.65 km

Therefore, the horizontal range of the projectile is approximately 7.65 kilometers.

Horizontal problem:

u = 400 cos 30 forever
range = u t
where t is the time from the vertical problem

Vertical problem:
Hi = 300 meters
Vi = 400 sin 30 = 200 m/s
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
0 = 300 + 200 t - 4.9 t^2
solve that quadratic for t (the positive root, not the one that occurs inside the virtual cliff on the way up)
Use that time in the air in the horizontal problem to get range.