We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 35 ∘. The gravitational acceleration is g=10 m/s2. (See figure)

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

To answer these questions, we can break down the motion of the stone into its horizontal and vertical components.

(a) To find the horizontal distance from the house wall where the stone hits the roof, we need to determine the time it takes for the stone to reach the roof and then calculate its horizontal displacement.

The initial vertical velocity of the stone can be found by multiplying the initial speed (v0) by the sine of the launch angle α:

vy0 = v0 * sin(α)
= 20 * sin(35°)
≈ 11.50 m/s

Using the equation of motion in the vertical direction, we can find the time it takes for the stone to reach the roof:

h = vy0 * t + (1/2) * g * t^2

Since we know the height of the house wall (h = 6 m), we can rearrange the equation to solve for time (t):

t^2 + 2 * vy0 * t / g - 2 * h / g = 0

Using the quadratic formula and considering only the positive solution:

t = (-2 * vy0 / g) + sqrt((2 * vy0 / g)^2 + 4 * (2 * h / g))
t ≈ 1.02 s

Once we have calculated the time, we can determine the horizontal displacement using the horizontal component of the initial velocity (v0x) and the time (t):

v0x = v0 * cos(α)
= 20 * cos(35°)
≈ 16.40 m/s

s = v0x * t
= 16.40 * 1.02
≈ 16.73 m

Therefore, the stone is going to hit the roof at a horizontal distance of approximately 16.73 meters from the house wall.

(b) We have already calculated the time it takes for the stone to reach the roof, which is approximately 1.02 seconds.