A skateboarder in a death defying stunt decides to launch herself from a ramp on a hill. The skateboarder leaves the ramp at a height of 1.4m above the slope, traveling 15m/s and at an angle of 40 degrees to the horizontal. The slope is inclined at 45 degrees to the horizontal. a). How far down the slope does the skateboarder land? b). How long is the skateboarder in the air? c). With what velocity does the skateboarder land on the slope?

To find the answers to these questions, we can break down the motion of the skateboarder into horizontal and vertical components. Let's solve each part step by step:

a) To find how far down the slope the skateboarder lands, we need to determine the horizontal distance traveled. We can use the horizontal component of the initial velocity.

The initial velocity can be found using the given speed of 15 m/s and the angle of 40 degrees. The horizontal velocity can be found by multiplying the initial velocity by the cosine of the angle:

horizontal velocity (Vx) = initial velocity (Vi) * cos(angle)

Vx = 15 m/s * cos(40 degrees)

Now, we can find the time it takes to travel down the slope using the horizontal distance and horizontal velocity.

horizontal distance = horizontal velocity * time

Using the distance formula, we can rearrange it to solve for time:

time = horizontal distance / horizontal velocity

b) To find the time the skateboarder is in the air, we can use the vertical motion of the skateboarder. The skateboarder starts at a height of 1.4 m above the slope.

To find the time of flight, we use the vertical component of the initial velocity. The vertical velocity can be found by multiplying the initial velocity by the sine of the angle:

vertical velocity (Vy) = initial velocity (Vi) * sin(angle)

Vy = 15 m/s * sin(40 degrees)

Now, we can find the time by using the equation for vertical motion:

height = vertical velocity * time - 0.5 * acceleration * time^2

At the highest point, the height will be 0, so we can set the equation equal to 0 and solve for time.

c) To find the velocity at which the skateboarder lands on the slope, we need to find the resultant velocity. We can use the horizontal and vertical components to find the resultant velocity using the Pythagorean theorem.

resultant velocity (Vr) = √(Vx^2 + Vy^2)

Now that we know the resultant velocity, we can determine the direction of the velocity using trigonometry. The angle can be found using the arctan function:

angle = arctan(Vy / Vx)

Now that we have the angle and magnitude of the velocity, we have all the information to answer the questions.

Let's calculate the values and find the answers to the questions.

To solve this problem, we can break it down into three parts: horizontal motion, vertical motion, and the combination of both.

a) To find how far down the slope the skateboarder lands, we need to find the horizontal displacement. We can use the horizontal component of the skateboarder's initial velocity, which is given by Vx_i = V_initial * cos(angle).

Let's substitute the values into the formula:
Vx_i = 15 m/s * cos(40 degrees)
Vx_i = 15 m/s * 0.766
Vx_i = 11.49 m/s

The horizontal displacement, Dx, can now be found using the formula:
Dx = Vx_i * t (where t is the time of flight)

However, we need to find the time of flight first.

b) To find the time of flight, we can use the vertical component of the skateboarder's initial velocity, which is given by Vy_i = V_initial * sin(angle).

Let's substitute the values into the formula:
Vy_i = 15 m/s * sin(40 degrees)
Vy_i = 15 m/s * 0.643
Vy_i = 9.645 m/s

To find the time of flight, t, we can use the equation:
Vy_f = Vy_i + g * t
where Vy_f is the final vertical velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the skateboarder lands on the slope (at the same height as the starting point), the final vertical velocity will be zero, so the equation becomes:
0 = 9.645 m/s - 9.8 m/s^2 * t

Solving for t:
9.8 m/s^2 * t = 9.645 m/s
t = 9.645 m/s / 9.8 m/s^2
t ≈ 0.9843 s

Now, we can find the horizontal displacement:
Dx = Vx_i * t
Dx = 11.49 m/s * 0.9843 s
Dx ≈ 11.31 m

Therefore, the skateboarder lands approximately 11.31 meters down the slope.

c) To find the velocity with which the skateboarder lands on the slope, we can use the equation for the resultant velocity, V_resultant.

V_resultant = sqrt(V_horizontal^2 + V_vertical^2)
where V_horizontal is the horizontal component of the velocity and V_vertical is the vertical component of the velocity.

Let's substitute the values into the formula:
V_resultant = sqrt((Vx_i)^2 + (Vy_i)^2)
V_resultant = sqrt((11.49 m/s)^2 + (9.645 m/s)^2)
V_resultant ≈ 15.05 m/s

Therefore, the skateboarder lands on the slope with a velocity of approximately 15.05 m/s.