A projectile is launched at an angle of 58.5 degrees above the horizontal and lands at the same level from which it was launched 3.88 seconds later. what is the magnitude of the initial velocity?

To find the magnitude of the initial velocity of the projectile, we can use the following kinematic equations of projectile motion:

1. Horizontal equation: \(x = v_{0x} \cdot t\)
2. Vertical equation: \(y = v_{0y} \cdot t - \frac{1}{2} g \cdot t^2\)
3. Velocity equation: \(v = \sqrt{{v_x}^2 + {v_y}^2}\)

Given:
- Launch angle: 58.5 degrees above horizontal
- Time of flight: 3.88 seconds

Let's start by breaking the initial velocity into its horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components.

The horizontal component of the initial velocity (\(v_{0x}\)) remains constant throughout the projectile's motion and can be calculated using \(v_{0x} = v \cdot \cos(\theta)\), where \(v\) is the magnitude of the initial velocity and \(\theta\) is the launch angle.

The vertical component of the initial velocity (\(v_{0y}\)) experiences acceleration due to gravity. At the highest point of the projectile's trajectory, the vertical velocity component is zero. Therefore, we can use the equation \(t = \frac{{v_{0y}}}{{g}}\) to find \(v_{0y}\).

Now, let's substitute the known values into the equations:

1. Horizontal equation: \(x = v_{0x} \cdot t\)
Since the projectile lands at the same level as its launch point, the horizontal displacement (\(x\)) is zero. So, \(0 = v_{0} \cdot \cos(\theta) \cdot t\) (1)

2. Vertical equation: \(y = v_{0y} \cdot t - \frac{1}{2} g \cdot t^2\)
Again, since the projectile lands at the same level as its launch point, the vertical displacement (\(y\)) is zero. So, \(0 = v_{0} \cdot \sin(\theta) \cdot t - \frac{1}{2} g \cdot t^2\) (2)

3. Velocity equation: \(v = \sqrt{{v_x}^2 + {v_y}^2}\)
The magnitude of the initial velocity (\(v\)) can be expressed as \(v = \sqrt{{v_{0x}}^2 + {v_{0y}}^2}\)

Now, we have two equations (equation 1 and equation 2) with two unknowns (\(v\) and \(v_{0y}\)). We can solve these equations simultaneously to find the values.

Substituting equation 1 into equation 2, we get:
\(0 = (v \cdot \sin(\theta)) \cdot t - \frac{1}{2} g \cdot t^2\)

Rearranging the equation, we have:
\(v = \frac{\frac{1}{2} g \cdot t^2}{\sin(\theta) \cdot t}\)

Now, substituting the given values:
\(v \approx \frac{\frac{1}{2} \cdot 9.8 \,\text{m/s}^2 \cdot (3.88 \,\text{s})^2}{\sin(58.5^\circ) \cdot 3.88 \,\text{s}}\)

Evaluating the expression will give you the magnitude of the initial velocity.