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1 month ago

What is the magnitude of the relative angle φ

Physics

1 answer:

kicyunya [3.2K]1 month ago

6 0

The complete question is;

A ski jumper descends a ramp and exits the ski track at a horizontal speed of 24 m/s. The slope at the landing site angles downwards at θ = 59◦. The acceleration due to gravity is 9.8 m/s².

What is the value of the relative angle φ at which the ski jumper makes contact with the slope? Provide the answer in degrees.

Answer:

14.08°

Explanation:

The time taken can be calculated using the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Next, the slope of the trajectory at the impact point is determined by;

tan α = V_y/V_x

Given V_x = 24 m/s

We will find V_y using;

v = gt

Therefore;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Consequently;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

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A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s

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