From the edge of the rooftop of a building, a boy throws a stone at an angle 26.1° above the horizontal. The stone hits the ground 4.25 s later, 103m away from the base of the building

(b) Find the initial velocity of the stone.

(c) Find the initial height h from which the stone was thrown.

I was able to figure out my missing equation. Please void this question

To find the initial velocity of the stone, you can use the horizontal motion equation. The horizontal distance is given as 103 m, and the time of flight is given as 4.25 s.

The horizontal motion equation is given by:
Distance = Velocity * Time

Substituting the given values:
103 m = Velocity * 4.25 s

To find the initial velocity, rearrange the equation:
Velocity = Distance / Time

Therefore, the initial velocity of the stone is:
Velocity = 103 m / 4.25 s = 24.24 m/s

To find the initial height from which the stone was thrown, you can use the vertical motion equation. The vertical distance is equal to the height of the building (h). The time of flight is given as 4.25 s, and the angle above the horizontal is given as 26.1°.

The vertical motion equation for an object in free fall is given by:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time^2

Since the stone is thrown horizontally, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2 (assuming no air resistance).

Substituting the given values:
h = 0 * 4.25 + 0.5 * 9.8 m/s^2 * (4.25 s)^2

Simplifying the equation:
h = 0 + 0.5 * 9.8 m/s^2 * (18.0625 s^2)

h = 0 + 88.355 m

Therefore, the initial height from which the stone was thrown is approximately 88.355 m.

To find the initial velocity of the stone, we can use the equation of motion for projectile motion:

R = v₀ * cos(θ) * t

where:
R is the horizontal distance traveled by the stone (103 m),
v₀ is the initial velocity of the stone,
θ is the angle above the horizontal (26.1°),
and t is the time of flight (4.25 s).

To solve for v₀, we need to rearrange the equation:

v₀ = R / (cos(θ) * t)

Now let's substitute the given values into the equation:

v₀ = 103 m / (cos(26.1°) * 4.25 s)

Now we can calculate the initial velocity of the stone:

v₀ ≈ 103 m / (0.8939 * 4.25 s)
v₀ ≈ 103 m / 3.797 s
v₀ ≈ 27.12 m/s

So, the initial velocity of the stone is approximately 27.12 m/s.

To find the initial height from which the stone was thrown, we can use the equation of motion for the vertical component of projectile motion:

h = v₀ * sin(θ) * t - (1/2) * g * t^2

where:
h is the initial height,
g is the acceleration due to gravity (approximately 9.8 m/s²).

Rearranging the equation to solve for h:

h = v₀ * sin(θ) * t - (1/2) * g * t^2

Let's substitute the known values:

h = 27.12 m/s * sin(26.1°) * 4.25 s - (1/2) * 9.8 m/s² * (4.25 s)²

Calculating:

h ≈ 27.12 m/s * sin(26.1°) * 4.25 s - (1/2) * 9.8 m/s² * 18.0625 s²

h ≈ 15.06 m - 88.445625 m

h ≈ -73.38 m

From the calculations, we find that the initial height from which the stone was thrown is approximately -73.38 m. The negative sign indicates that the stone was likely thrown from a point below the reference level (the ground). It is important to double-check the signs and assumptions made during the calculations, as negative values for height may not be physically meaningful in this context.