A projectile is shot from the ground at an angle of 60 degrees with respect to the horizontal, and it lands on the ground 5 seconds later. Find:

a. the horizontal component of initial velocity
b. the vertical component of initial velocity
c. initial speed

Let Vyo be the initial vertical velocity component. It equals g times half the time of flight. Lex Vx be the horizontal velocity component, which is constant while in flight.

b. (1/2)*g*5 sec = Vyo = 24.5 m/s
a. Vyo/Vx = tan 60 = 1.732
Vx = 14.14 m/s
c. Vo = sqrt[(Vx^2 + Vyo^2]

To find the horizontal and vertical components of the initial velocity, as well as the initial speed, we can use the following formulas:

1. Horizontal component of initial velocity (Vx):
Vx = V0 * cosθ

2. Vertical component of initial velocity (Vy):
Vy = V0 * sinθ

3. Initial speed (V0):
V0 = sqrt(Vx² + Vy²)

Here, V0 represents the initial velocity, θ represents the launch angle, Vx represents the horizontal component of the initial velocity, and Vy represents the vertical component of the initial velocity.

Given:

Launch angle (θ) = 60 degrees
Time of flight (t) = 5 seconds

Step 1: Finding Vx
Using formula (1), we can calculate the horizontal component of the initial velocity (Vx):

Vx = V0 * cosθ

Since we don't have the value of V0, let's first find V0 using formula (3) to substitute it back into formula (1).

Step 2: Finding Vy
Using formula (2), we can calculate the vertical component of the initial velocity (Vy):

Vy = V0 * sinθ

Step 3: Finding V0
Using formula (3), we can calculate the initial speed (V0):

V0 = sqrt(Vx² + Vy²)

Now let's solve these equations one by one.

Step 1:
Vx = V0 * cosθ

Since the launch angle θ = 60 degrees, we have:
Vx = V0 * cos60

Step 2:
Vy = V0 * sinθ

Using the same launch angle value as before (θ = 60 degrees), we have:
Vy = V0 * sin60

Step 3:
V0 = sqrt(Vx² + Vy²)

Substituting Vx and Vy into this equation:
V0 = sqrt((V0 * cos60)² + (V0 * sin60)²)

This equation involves a variable V0 on both sides. To solve it, we need to simplify it algebraically.

Multiplying out both sides of the equation:
V0 = sqrt(V0² * cos²60 + V0² * sin²60)

Since cos²60 = 0.5 and sin²60 = 0.75, the equation simplifies to:
V0 = sqrt(V0² * 0.5 + V0² * 0.75)

Combining like terms:
V0 = sqrt(1.25 * V0²)

Squaring both sides to eliminate the square root:
V0² = 1.25 * V0²

Now, isolate V0:
V0² - 1.25 * V0² = 0
-0.25 * V0² = 0
V0² = 0 / (-0.25)
V0² = 0

Taking the square root of both sides:
V0 = sqrt(0)
V0 = 0

Since V0 is the initial speed, we have found that the initial speed is 0.

Therefore, in this case, the projectile was not launched with any initial speed, and it did not defy gravity to stay in the air. Please check the given information and calculations to ensure accuracy.