A ski jumper starts from rest 44.0 m above the ground on a frictionless track, and flies off the track at an angle of 45.0° above the horizontal and at a height of 14.0 m from the ground. Neglect air resistance.

(a) What is his speed when he leaves the track?


m/s
(b) What is the maximum altitude he attains after leaving the track?


m
(c) Where does he land relative to the end of the track?
m

To solve this problem, we can use the principles of conservation of energy and projectile motion.

(a) To find the speed when the ski jumper leaves the track, we can use the principle of conservation of energy. The total initial mechanical energy is equal to the total final mechanical energy. Let's assume the ground level as the reference point for potential energy.

Initial potential energy (mgh1) + Initial kinetic energy (0) = Final potential energy (mgh2) + Final kinetic energy (1/2mv^2)

Given:
Initial height h1 = 44.0 m
Final height h2 = 14.0 m
Angle above the horizontal θ = 45.0°
Acceleration due to gravity g = 9.8 m/s^2

Using the equation for potential energy and simplifying:
mgh1 = mgh2 + (1/2)mv^2

Canceling out the mass, we get:
gh1 = gh2 + (1/2)v^2

Simplifying further:
gh1 - gh2 = (1/2)v^2
g(h1 - h2) = (1/2)v^2

Substituting the given values:
(9.8)(44.0 - 14.0) = (1/2)v^2

Solving for v:
(9.8)(30.0) = (1/2)v^2
294 = (1/2)v^2
588 = v^2
v = √588 ≈ 24.2 m/s

Therefore, the speed when the ski jumper leaves the track is approximately 24.2 m/s.

(b) To find the maximum altitude the ski jumper attains after leaving the track, we can use the equation for projectile motion. The vertical component of velocity at the maximum altitude is zero.

Vertical component of initial velocity Vy = v * sin(θ)
Vertical component of final velocity Vy = 0

Using the equation for vertical motion:
Vy^2 = Vfy^2 - 2gΔy

Since Vy = 0, the equation simplifies to:
0 = (0)^2 - 2gΔy

Simplifying further:
2gΔy = 0

Δy = 0

Therefore, the maximum altitude the ski jumper attains after leaving the track is the same as the initial height, which is 44.0 m.

(c) Finally, to determine where the ski jumper lands relative to the end of the track, we need to find the horizontal distance traveled. This can be done using the equation for horizontal motion.

Horizontal component of initial velocity Vx = v * cos(θ)

Using the equation for horizontal motion:
Vx = dx / t
dx = Vx * t

Since the horizontal component of velocity remains constant throughout the motion, we only need to find the time of flight.

Using the equation for vertical motion:
Vy = Voy + gt
0 = v * sin(θ) - gt

Solving for t:
t = v * sin(θ) / g

Substituting the given values:
t = (24.2 m/s) * sin(45.0°) / 9.8 m/s^2
t ≈ 1.56 s

Now, calculating the horizontal distance traveled:
dx = (24.2 m/s) * cos(45.0°) * 1.56 s
dx ≈ 24.2 m

Therefore, the ski jumper lands approximately 24.2 m beyond the end of the track.

To find the answers to the given questions, we can use the principles of energy conservation and projectile motion. Let's break down the problem step by step:

(a) To determine the speed when the ski jumper leaves the track, we need to use the principle of conservation of mechanical energy. The initial mechanical energy (at the starting point) is equal to the final mechanical energy (when leaving the track). Since the ski jumper starts from rest, they have zero kinetic energy initially.

The initial mechanical energy is given by the potential energy, which is mgh (mass x gravitational acceleration x height). Once the ski jumper leaves the track, the only remaining energy at that point is the kinetic energy, which is given by 1/2 * m * v² (mass x velocity squared).

Setting the initial mechanical energy equal to the final kinetic energy, we can solve for the velocity:

Potential energy at start = Kinetic energy when leaving track
mgh = 1/2 * m * v²

The mass cancels out, and we can solve for v:

v = sqrt(2 * g * h)

Here, g represents the acceleration due to gravity (9.8 m/s²) and h is the height above the ground (44.0 m). Plugging in the values:

v = sqrt(2 * 9.8 * 44.0)
v = 29.84 m/s (rounded to two decimal places)

So, the speed when the ski jumper leaves the track is approximately 29.84 m/s.

(b) To determine the maximum altitude the ski jumper attains after leaving the track, we need to analyze the projectile motion. The horizontal and vertical motions are independent of each other.

First, we can determine how long the ski jumper stays in the air. We can use the fact that the time of flight for a projectile is the same for upward and downward motion. The formula for the time of flight is given by:

T = 2 * (v * sin(angle) / g)

Here, v is the initial velocity (29.84 m/s), the angle is 45 degrees, and g is the acceleration due to gravity (9.8 m/s²). Plugging in the values:

T = 2 * (29.84 * sin(45) / 9.8)
T = 4.29 s (rounded to two decimal places)

Next, we can calculate the maximum height attained by using the time of flight. The formula for the maximum height of a projectile launched at an angle is:

h_max = (v^2 * sin^2(angle)) / (2 * g)

Plugging in the values:

h_max = (29.84^2 * sin^2(45)) / (2 * 9.8)
h_max = 29.05 m (rounded to two decimal places)

Therefore, the maximum altitude the ski jumper attains after leaving the track is approximately 29.05 m.

(c) To determine where the ski jumper lands relative to the end of the track, we can consider the horizontal distance traveled during the flight. Since there is no horizontal acceleration (neglecting air resistance), the horizontal distance traveled, x, can be found using the formula:

x = v * cos(angle) * T

Here, v is the initial velocity (29.84 m/s), the angle is 45 degrees, and T is the time of flight (4.29 s). Plugging in the values:

x = 29.84 * cos(45) * 4.29
x = 109.77 m (rounded to two decimal places)

Therefore, the ski jumper lands approximately 109.77 m from the end of the track.

4.0