From the edge of the rooftop of a building, a boy throws a stone at an angle 30.0° above the horizontal. The stone hits the ground 4.10 s later, 112 m away from the base of the building. (Ignore air resistance.)

(a) Find the initial velocity of the stone.
magnitude _____
direction_________

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

(c) Find the maximum height H reached by the stone.

To solve this problem, we can use the equations of motion for projectile motion.

Let's first analyze the horizontal motion of the stone. The horizontal velocity remains constant throughout the motion because there is no horizontal acceleration. The horizontal distance traveled by the stone is given as 112 m, and the time taken is 4.10 s. Therefore, we can use the formula:

Horizontal distance = Horizontal velocity × Time

Rearranging the equation, we can find the horizontal velocity:

Horizontal velocity = Horizontal distance / Time = 112 m / 4.10 s

Solving the equation, we get:

Horizontal velocity = 27.32 m/s

For part (a), we need to find the initial velocity of the stone. Since the angle of projection is given as 30.0° above the horizontal, we can represent the initial velocity as the sum of horizontal and vertical components.

The vertical component of velocity can be found using the formula:

Vertical velocity = Initial velocity × sin(θ)

Here, θ is the angle of projection (30.0°) and the vertical velocity component is opposite to gravity. We know that the time taken for the stone to hit the ground is 4.10 s, and we can use this time to find the vertical component of velocity.

The equation for vertical motion is:

Vertical distance = Initial vertical velocity × Time - (1/2) × g × Time^2

Rearranging the equation, we can find the initial vertical velocity:

Initial vertical velocity = (Vertical distance + (1/2) × g × Time^2) / Time

Given that the stone hit the ground, the vertical distance traveled is equal to zero. Therefore, the equation becomes:

Initial vertical velocity = (1/2) × g × Time

Solving the equation, we get:

Initial vertical velocity = 0.5 × 9.8 m/s^2 × 4.10 s = 20.09 m/s

Now, we can find the initial velocity of the stone using the Pythagorean theorem:

Initial velocity = √[(Horizontal velocity)^2 + (Vertical velocity)^2]

Substituting the values, we get:

Initial velocity = √[(27.32 m/s)^2 + (20.09 m/s)^2]

Solving the equation, we find:

Initial velocity = 33.40 m/s

For part (b), we can calculate the initial height (h) from which the stone was thrown.

The equation for vertical motion is:

Vertical distance = Initial vertical velocity × Time - (1/2) × g × Time^2

We know that the time taken is 4.10 s, and the vertical distance traveled is equal to zero at the maximum height.

Therefore, the equation becomes:

h = (1/2) × g × Time^2

Substituting the values, we get:

h = (1/2) × 9.8 m/s^2 × (4.10 s)^2

Solving the equation, we find:

h = 84.50 m

For part (c), we need to calculate the maximum height (H) reached by the stone.

The equation for vertical motion is:

Vertical distance = Initial vertical velocity × Time - (1/2) × g × Time^2

We know that at the maximum height, the vertical velocity is equal to zero. Therefore, the equation becomes:

H = Initial vertical velocity × Time

Substituting the values, we get:

H = 20.09 m/s × 4.10 s

Solving the equation, we find:

H = 82.29 m

To summarize:

(a) The initial velocity of the stone is 33.40 m/s.
(b) The initial height from which the stone was thrown is 84.50 m.
(c) The maximum height reached by the stone is 82.29 m.