A ski jumper travels down a slope and leaves
the ski track moving in the horizontal direction with a speed of 21 m/s as in the figure.
The landing incline below her falls off with a
slope of θ = 49◦
.
The acceleration of gravity is 9.8 m/s
2
Calculate the distance d she travels along
the incline before landing.
Answer in units of m.
To calculate the distance the ski jumper travels along the incline before landing, we can use the principles of projectile motion.
First, let's break down the initial velocity of the ski jumper into its horizontal and vertical components. Given that the horizontal speed is 21 m/s and the slope has an angle of θ = 49°, we can calculate the horizontal component of velocity using trigonometry:
Horizontal component of velocity (Vx) = speed * cos(θ)
Vx = 21 m/s * cos(49°)
Next, let's find the vertical component of velocity using the same trigonometry:
Vertical component of velocity (Vy) = speed * sin(θ)
Vy = 21 m/s * sin(49°)
Now, since we are interested in calculating the distance traveled along the incline, we only need the horizontal component of velocity. The initial horizontal velocity will remain constant throughout the motion, while the vertical velocity will be affected by gravity.
The time it takes for the ski jumper to land can be calculated using the vertical motion equation:
vf = vi + at
Where vf is the final vertical velocity (when the ski jumper lands), vi is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
When the ski jumper lands, its vertical velocity (Vy) will become zero. Therefore:
0 = Vy + (-9.8 m/s^2) * t
Solving for t:
t = Vy / 9.8 m/s^2
Now let's find the horizontal distance traveled, which is the distance along the incline:
Distance (d) = Vx * t
Substituting the values we have:
d = (21 m/s * cos(49°)) * (Vy / 9.8 m/s^2)
Now we can calculate the value of d by plugging in the calculated values of Vx and Vy:
d = (21 m/s * cos(49°)) * (21 m/s * sin(49°) / 9.8 m/s^2)
Calculating this expression will give us the distance d traveled along the incline before landing.
To solve this problem, we can use the principles of projectile motion.
First, we can break down the initial velocity of the ski jumper into its horizontal and vertical components. Since the ski jumper leaves the ski track moving horizontally with a speed of 21 m/s, the initial horizontal velocity (Vx) is 21 m/s. The initial vertical velocity (Vy) is 0 m/s because there is no vertical component of the initial velocity.
Next, we need to find the time it takes for the ski jumper to land. We can use the equation of motion in the vertical direction:
h = (1/2)gt^2
where h is the vertical distance traveled and g is the acceleration due to gravity. Since the ski jumper is landing on an incline, we can consider the height h as the vertical distance from the point where the ski jumper leaves the ski track to the point where she lands on the incline.
The vertical distance, h, can be calculated using the trigonometric relationship:
h = x * tan(θ)
where x is the horizontal distance traveled along the incline and θ is the angle of the incline. Given that θ = 49°, we can solve for h:
h = x * tan(49°)
Now, we can substitute this expression for h into the equation of motion:
x * tan(49°) = (1/2)gt^2
Since the time t it takes for the ski jumper to land is the same for the horizontal and vertical motion, we can solve for t:
x * tan(49°) = (1/2)g * t^2
t^2 = (2 * x * tan(49°)) / g
Taking the square root of both sides, we get:
t = sqrt((2 * x * tan(49°)) / g)
Next, we can use the equation of motion in the horizontal direction:
x = Vx * t
Substituting the expression for t, we get:
x = Vx * sqrt((2 * x * tan(49°)) / g)
Now we can solve for x:
x^2 = (Vx^2 * 2 * x * tan(49°)) / g
x^2 * g = Vx^2 * 2 * x * tan(49°)
x * g = Vx^2 * 2 * tan(49°)
Simplifying further, we get:
x = (Vx^2 * 2 * tan(49°)) / g
Now we can substitute the given values into the equation to find the distance x:
x = (21^2 * 2 * tan(49°)) / 9.8
x ≈ 115.97 m
Therefore, the ski jumper travels approximately 115.97 m along the incline before landing.
To calculate the distance the ski jumper travels along the incline before landing, we can use the equations of motion.
First, we need to find the time it takes for the ski jumper to land. We can use the equation for horizontal motion:
distance = speed x time
Since the horizontal speed is given as 21 m/s and the distance is not provided, we can't directly find the time just yet.
However, we can find the vertical component of the ski jumper's initial velocity using trigonometry. The vertical speed can be calculated using the angle θ:
vertical speed = initial speed x sin(θ)
In this case, the initial speed is also the horizontal speed, which is 21 m/s, and θ is 49°. Substituting these values:
vertical speed = 21 m/s x sin(49°)
Now, let's calculate the time it takes for the ski jumper to land using the vertical motion equation:
distance = initial velocity x time + (0.5 x acceleration x time^2)
The initial velocity in the vertical direction is the calculated vertical speed, the distance is unknown (to be found), the acceleration is due to gravity (-9.8 m/s^2), and we solve for time.
Since the ski jumper lands on the inclined slope, the vertical distance traveled is zero when she lands. We can substitute these values into the equation:
0 = vertical speed x time + (0.5 x acceleration x time^2)
Rearranging the equation gives us a quadratic equation:
0.5 x (-9.8 m/s^2) x time^2 + vertical speed x time = 0
Using the quadratic formula, we can solve for the time it takes for the ski jumper to land:
time = (-vertical speed ± √(vertical speed^2 - 4 x 0.5 x (-9.8 m/s^2) x 0)) / 2 x 0.5 x (-9.8 m/s^2)
Since the time cannot be negative, we only consider the positive root:
time = (-vertical speed + √(vertical speed^2)) / (-9.8 m/s^2)
Now that we have the time, we can calculate the distance traveled along the incline before landing using the horizontal motion equation:
distance = speed x time
Substituting the values:
distance = 21 m/s x time
Finally, we can calculate the distance d she travels along the incline before landing:
distance = 21 m/s x [( -vertical speed + √(vertical speed^2)) / (-9.8 m/s^2)]
Note: Remember to convert the angle θ to radians when taking the sine.
Hope this helps!