A projectile that is launched at 15 degrees and the landing height is the same as the launch position. If this is true, then at what other angle can the projectile be launch to achieve the same horizontal displacement?

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To find the angle at which the projectile can be launched to achieve the same horizontal displacement, we need to understand the motion of a projectile.

When a projectile is launched, it follows a curved path due to its initial velocity and the force of gravity acting on it. The path of a projectile can be broken down into two independent components: horizontal and vertical motion.

In the given scenario, the projectile is launched at an angle of 15 degrees and the landing height is the same as the launch position. This means that the vertical displacement of the projectile is zero. To achieve the same horizontal displacement, we need to find the angle at which the horizontal displacement remains constant.

To solve this problem, we can use the equations of motion for projectile motion. The horizontal and vertical components of velocity can be represented as:

Vx = Vo * cos(theta)
Vy = Vo * sin(theta)

Where Vx is the horizontal component of velocity, Vy is the vertical component of velocity, Vo is the initial velocity, and theta is the launch angle.

Since the projectile lands at the same position, the final vertical displacement is also zero. We can use this information to find the time of flight (t) using the equation:

0 = (Vy * t) + (0.5 * g * t^2)

Since the vertical displacement is zero, this equation simplifies to:

0 = (Vo * sin(theta) * t) - (0.5 * g * t^2)

Simplifying further, we get:

t * (Vo * sin(theta) - 0.5 * g * t) = 0

This equation has two solutions: t = 0 (which corresponds to the launch position) and (Vo * sin(theta) - 0.5 * g * t) = 0.

To find the angle at which the projectile can be launched to achieve the same horizontal displacement, we need to solve for theta from the second equation. Let's rearrange it:

Vo * sin(theta) - 0.5 * g * t = 0

Now, substitute the value of t from the equation:

(Vo * sin(theta)) - 0.5 * g * [(2 * Vo * sin(theta)) / g] = 0

Simplifying further:

Vo * sin(theta) - Vo * sin(theta) = 0

This equation cancels out Vo * sin(theta) on both sides, leaving us with 0 = 0. This means that for any value of theta, the equation holds true.

Therefore, any angle of launch will result in the same horizontal displacement when the landing height is the same as the launch position.