A hammer slides down a roof of angle theta (with respect to the ground). It slides along the

roof distance D. As it leaves the roof, a height H above the ground, it has velocity in both
directions. Find how far from the base of the building the hammer lands.

To find how far from the base of the building the hammer lands, we can break down the motion into horizontal and vertical components.

Let's denote:
- θ as the angle of the roof with respect to the ground,
- D as the horizontal distance the hammer slides along the roof, and
- H as the vertical height of the hammer above the ground.

First, let's find the initial velocity of the hammer just before it leaves the roof. We can use the vertical component of the velocity to do this.

The vertical velocity component can be calculated using the equation:

V_y = V * sin(θ)

where V is the magnitude of the velocity vector (which will be the same for both horizontal and vertical components).

Since the hammer leaves the roof with a velocity in both directions, the vertical velocity component will be positive √2 times the magnitude:

V_y = √2 * V * sin(θ)

Next, we can find the time it takes for the hammer to reach the ground by considering the vertical motion alone.

The vertical displacement is given by the equation:

H = V_y * t - (1/2) * g * t^2

where g is the acceleration due to gravity and t is the time.

Since the hammer is at its maximum height when it leaves the roof, the vertical displacement H will be 0.

Therefore, we can set up the equation:

0 = √2 * V * sin(θ) * t - (1/2) * g * t^2

Rearranging the equation, we get:

(1/2) * g * t^2 = √2 * V * sin(θ) * t

Simplifying, we have:

g * t = 2√2 * V * sin(θ)

Now, we can find the time t:

t = (2√2 * V * sin(θ)) / g

Finally, we can find the horizontal distance the hammer lands from the base of the building. This is given by the equation:

Range = D + (V * cos(θ)) * t

Substituting the expression for t, we get:

Range = D + (V * cos(θ)) * (2√2 * V * sin(θ)) / g

Simplifying further, we have:

Range = D + (2√2 * V^2 * cos(θ) * sin(θ)) / g

Therefore, the hammer will land a distance of D + (2√2 * V^2 * cos(θ) * sin(θ)) / g from the base of the building.

To find how far from the base of the building the hammer lands, we can break down the problem into horizontal and vertical components.

Let's start by analyzing the vertical motion of the hammer. We can use the equations of motion to determine the time of flight and the final vertical velocity when the hammer hits the ground.

In the vertical direction, the initial velocity of the hammer is zero (since it only slides down the roof). We know that the displacement in the vertical direction is equal to the height H above the ground. We can use the equation of motion:

H = (1/2) * g * t^2 (Equation 1)

where g is the acceleration due to gravity and t is the time of flight.

We can also determine the final vertical velocity using the equation of motion:

v_fy = g * t (Equation 2)

Now, let's analyze the horizontal motion. The horizontal component of the velocity remains constant throughout the motion since there are no horizontal forces acting on the hammer. The horizontal displacement, which is the distance from the base of the building where the hammer lands, can be determined using the equation:

D = v_fx * t (Equation 3)

where v_fx is the horizontal component of the velocity and t is the time of flight.

To find the horizontal component of the velocity, we need to resolve the initial velocity into its horizontal and vertical components. The horizontal component is given by:

v_ix = v_i * cos(theta) (Equation 4)

where v_i is the initial velocity of the hammer and theta is the angle of the roof with respect to the ground.

Now, we have the necessary equations to solve for the time of flight, the final vertical velocity, and the horizontal displacement:

1. Rearrange Equation 1 to solve for t:
t = sqrt((2H) / g)

2. Substitute the value of t into Equation 2 to find the final vertical velocity:
v_fy = g * sqrt((2H) / g)

3. Substitute the value of t into Equation 3 to find the horizontal displacement:
D = v_fx * sqrt((2H) / g)

4. Substitute the value of v_fx from Equation 4 into Equation 3:
D = (v_i * cos(theta)) * sqrt((2H) / g)

By using these equations and plugging in the given values for H, v_i, theta, and g, you can find the horizontal displacement D—the distance from the base of the building where the hammer lands.