A ski jumper starts from rest 51.0 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 12.0 m above the ground. Neglect air resistance.

(a) What is her speed when she leaves the track?

(b) What is the maximum altitude she attains after leaving the track?

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

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

(a) To find the speed when she leaves the track, we can use the principle of conservation of mechanical energy. The initial mechanical energy is equal to the final mechanical energy.

Initial mechanical energy (Ei) = Potential energy at the starting point
Final mechanical energy (Ef) = Potential energy at the highest point + Kinetic energy at the highest point

Ei = Ef

The initial potential energy is given by mgh, where m is the mass of the ski jumper, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height above the ground.

Ei = mgh

The final potential energy at the highest point is given by mgh_f, where h_f is the maximum altitude she attains after leaving the track.

The final kinetic energy at the highest point is given by (1/2)mv^2, where v is the velocity (speed) when she leaves the track.

Ef = mgh_f + (1/2)mv^2

Setting Ei equal to Ef:

mgh = mgh_f + (1/2)mv^2

Canceling out the mass (m) from both sides of the equation:

gh = gh_f + (1/2)v^2

Rearranging the equation to solve for v:

(1/2)v^2 = gh - gh_f

v^2 = 2gh - 2gh_f

Taking the square root of both sides to solve for v:

v = √(2gh - 2gh_f)

Now we can plug in the given values:

g = 9.8 m/s^2
h = 51.0 m
h_f = 12.0 m

v = √(2 * 9.8 * 51 - 2 * 9.8 * 12)

Calculating the value of v gives:

v ≈ 29.4 m/s

Therefore, the speed when she leaves the track is approximately 29.4 m/s.

(b) To find the maximum altitude she attains after leaving the track, we can use the conservation of mechanical energy principle again. At the highest point, the kinetic energy is zero.

Ef = mgh_f + (1/2)mv^2 = 0

Rearranging the equation to solve for h_f:

mgh_f = - (1/2)mv^2

Canceling out the mass (m) from both sides of the equation:

gh_f = - (1/2)v^2

Solving for h_f:

h_f = - (1/2)v^2/g

Substituting the values:

v = 29.4 m/s
g = 9.8 m/s^2

h_f = - (1/2) * (29.4)^2 / 9.8

Calculating the value of h_f gives:

h_f ≈ -45.5 m

Since height cannot be negative, the maximum altitude she attains after leaving the track is approximately 45.5 m above the ground.

(c) To find where she lands relative to the end of the track, we can use the horizontal component of her motion. Since there is no horizontal force, her horizontal velocity remains constant throughout the flight.

The horizontal distance she covers can be calculated using the equation:

Range = Horizontal velocity * Time of flight

The time of flight can be calculated using the equation:

Time of flight = 2 * Vertical velocity / g

Vertical velocity at the highest point is zero, so the time of flight is equal to twice the time it takes her to reach the highest point.

t = 2 * √(2h/g)

Substituting the values:

h = 51.0 m
g = 9.8 m/s^2

t = 2 * √(2 * 51.0 / 9.8)

Calculating the value of t gives:

t ≈ 4.19 s

The horizontal velocity can be calculated using the equation:

v_horizontal = v * cos(45°)

Substituting the value of v:

v = 29.4 m/s

v_horizontal = 29.4 * cos(45°)

Calculating the value of v_horizontal gives:

v_horizontal ≈ 20.8 m/s

Finally, we can find the horizontal distance she covers:

Range = v_horizontal * t

Substituting the values:

v_horizontal = 20.8 m/s
t = 4.19 s

Range = 20.8 * 4.19

Calculating the value of Range gives:

Range ≈ 87.2 m

Therefore, she lands approximately 87.2 meters beyond the end of the track.

To find the answers to these questions, we can use the principles of conservation of energy and projectile motion.

Let's break down the problem step by step:

(a) To find the ski jumper's speed when she leaves the track, we can use the principle of conservation of energy. At the start, she only has gravitational potential energy, and at the highest point of her trajectory, she only has kinetic energy. Therefore, we can equate these two energies.

The gravitational potential energy at the start is given by:
PE(start) = m * g * h(start)

where m is the mass of the ski jumper, g is the acceleration due to gravity, and h(start) is the height above the ground at the start.

The kinetic energy at the highest point is given by:
KE(highest) = (1/2) * m * v(highest)^2

where v(highest) is the velocity at the highest point of the trajectory.

Since energy is conserved, we have:
PE(start) = KE(highest)

Substituting the values given in the problem, we get:
m * g * h(start) = (1/2) * m * v(highest)^2

Simplifying and solving for v(highest), we get:
v(highest) = sqrt(2 * g * h(start))

Substituting the given values into the equation, we can calculate the ski jumper's speed when she leaves the track.

(b) To find the maximum altitude she attains after leaving the track, we need to calculate the maximum height of her trajectory. Using projectile motion equations, we can find the maximum height when the vertical component of the velocity becomes zero.

The vertical component of the velocity at the highest point can be written as:
v_vertical(highest) = v(highest) * sin(45°)

Using kinematic equations for motion in the vertical direction, we have:
v_vertical(highest)^2 = v_initial^2 - 2 * g * h_max

At the highest point, v_vertical(highest) becomes zero. Substituting this into the equation, we get:
0 = v_initial^2 - 2 * g * h_max

Solving for h_max, we have:
h_max = (v_initial^2) / (2 * g)

Substituting the values given in the problem, we can calculate the maximum altitude she attains after leaving the track.

(c) To find where she lands relative to the end of the track, we can consider the horizontal motion of the ski jumper. Since there is no air resistance, the horizontal component of her velocity remains constant throughout the motion.

The horizontal component of her velocity can be written as:
v_horizontal = v(highest) * cos(45°)

Using the equation for motion in the horizontal direction, we have:
distance = v_horizontal * time

The time of flight can be calculated using the equation:
time = 2 * v_vertical(highest) / g

Substituting the values given in the problem, we can calculate the distance she lands relative to the end of the track.

By following these steps, you can find the answers to each part of the question. Remember to use the given values and equations accurately to obtain the correct results.

KE=PE

1/2mv^2=mgh