A Nordic jumper goes off a ski jump at an angle of 10.0° below the horizontal, traveling 92.0 m horizontally and 47.0 m vertically before landing.; (a) Ignoring friction and aerodynamic effects, calculate the speed needed by the skier on leaving the ramp
v(x) =x/t =>
t=x/v⒳=x/v₀•cosα
y=v₀⒴•t +gt²/2 =
= v₀⒴• x/v₀•cosα +g•x²/2v₀²•cos²α=
=x•tanα + g•x²/2v₀²•cos²α.
v₀=sqrt{g•x²/2•cos²α(y-x•tan α)}=
=sqrt{9.8•92²/2•0.97•(47-92•tan10)}=
=37.3 m/s.
A rowboat crosses a river with a velocity of 4.23 mi/h at an angle 62.5o north of west relative to the water. The river is 0.543 mi wide and carries an eastward current of 0.73 mi/h. How far upstream is the boat when it reaches the opposite shore?
To calculate the speed needed by the skier on leaving the ramp, we can use the principle of conservation of energy.
The initial potential energy of the skier on leaving the ramp will be equal to the final kinetic energy at the highest point of the jump.
The potential energy at the highest point of the jump is given by:
PE = mgh
Where m is the mass of the skier, g is the acceleration due to gravity, and h is the vertical distance traveled by the skier.
The kinetic energy at the highest point of the jump is given by:
KE = (1/2)mv^2
Where v is the velocity/speed of the skier.
Since the skier is initially at rest, the initial kinetic energy is zero. Therefore, the initial potential energy is equal to the final kinetic energy at the highest point of the jump.
So, mgh = (1/2)mv^2
In this case, we can consider the mass of the skier to be canceled out.
Therefore, gh = (1/2)v^2
Simplifying, we get:
v^2 = 2gh
Taking the square root of both sides, we get:
v = √(2gh)
In this case, the vertical distance traveled by the skier is 47.0 m. The acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the speed needed by the skier on leaving the ramp is:
v = √(2 * 9.8 * 47.0)
v ≈ √(921.6)
v ≈ 30.4 m/s
Hence, the speed needed by the skier on leaving the ramp is approximately 30.4 m/s.
To calculate the speed needed by the skier on leaving the ramp, we can use the principles of projectile motion.
Projectile motion can be separated into horizontal and vertical components. Since the skier leaves the ramp at an angle of 10.0° below the horizontal, we need to find the initial velocity of the skier's motion in both the horizontal and vertical directions.
Let's start by calculating the vertical component of the skier's motion:
Vertical motion:
1. Identify the known variables:
- Initial vertical position (y₀) = 0 m (skier starts from the ground)
- Final vertical position (y) = 47.0 m
- Initial vertical velocity (v₀y) = ?
- Final vertical velocity (v) = 0 m/s (skier reaches the highest point and descends)
- Acceleration due to gravity (g) = 9.8 m/s²
2. Apply the kinematic equation for vertical motion:
y = y₀ + v₀y * t + (1/2) * g * t²
3. Since the skier's initial vertical position is 0 m and the final vertical position is 47.0 m (going upwards), we can rearrange the equation as follows:
47.0 m = (1/2) * 9.8 m/s² * t²
4. Solve for the time (t):
t = sqrt((2 * 47.0 m) / 9.8 m/s²) ≈ 3.02 s
Now that we have the time taken to reach the highest point, we can calculate the initial velocity in the vertical direction using another kinematic equation:
v = v₀y + g * t
0 m/s = v₀y + 9.8 m/s² * 3.02 s
Solving for v₀y:
v₀y = -9.8 m/s² * 3.02 s = -29.60 m/s
Note: Negative sign indicates that the initial vertical velocity is in the downward direction.
Next, we can calculate the horizontal component of the skier's motion:
Horizontal motion:
1. Identify the known variables:
- Horizontal distance (x) = 92.0 m
- Initial horizontal velocity (v₀x) = ?
- Final horizontal velocity (v) = v₀x (horizontal velocity is constant as there is no horizontal acceleration)
2. Apply the equation for constant velocity:
x = v₀x * t
3. Rearrange the equation to solve for v₀x:
v₀x = x / t = 92.0 m / 3.02 s ≈ 30.46 m/s
Now we have both the initial vertical velocity (v₀y ≈ -29.60 m/s) and the initial horizontal velocity (v₀x ≈ 30.46 m/s).
To find the total initial speed (v₀), we can use the Pythagorean theorem:
v₀ = sqrt(v₀x² + v₀y²)
= sqrt((30.46 m/s)² + (-29.60 m/s)²)
≈ 41.05 m/s
Therefore, the speed needed by the skier on leaving the ramp (ignoring friction and aerodynamic effects) is approximately 41.05 m/s.