An extreme skier takes off from a jump and is going for height? if she leaves the jump traveling 13 m/s as she reaches the top of her flight, how high does she get above the top of the ramp?
Well, if she's an extreme skier, she must have some serious altitude aspirations! Let's crunch the numbers and find out how high she goes.
To figure out the height, we'll need a little physics magic. We know that the initial velocity (v₀) is 13 m/s, and at the top of the flight, the velocity (v) becomes zero (since she's at the peak).
We can use the kinematic equation that relates final velocity (v), initial velocity (v₀), acceleration (a), and displacement (s):
v² = v₀² + 2as
Since she reaches zero velocity at the top, we can simplify the equation to:
0 = (13 m/s)² + 2a(s)
Now, we just need to solve for s, which is the displacement or the height above the top of the ramp.
Don't panic, I'm about to do some magical calculations...
(13 m/s)² = 2a(s)
169 m²/s² = 2a(s)
Now, let's assume that the acceleration (a) due to gravity is approximately 9.8 m/s² (let's ignore air resistance because extreme skiers are known for defying gravity and the laws of physics!).
169 m²/s² = 2(9.8 m/s²)(s)
s ≈ 169 m²/s² / (2 * 9.8 m/s²)
s ≈ 8.63 m
So, she gets about 8.63 meters high above the top of the ramp, defying gravity like a skier superhero! Keep skiing, you gravity-defying daredevil!
To calculate the height the skier reaches above the top of the ramp, we can use the concept of projectile motion. The initial velocity of the skier can be broken down into its vertical and horizontal components.
Given:
Initial velocity (v₀) = 13 m/s
At the top of the flight, the vertical component of the velocity becomes zero (vₑ = 0). We can use the vertical motion equation:
vₑ² = v₀² - 2gΔy
Where:
vₑ = final vertical velocity (0 m/s)
v₀ = initial vertical velocity (13 m/s)
g = acceleration due to gravity (9.8 m/s²)
Δy = change in height
Rearranging the equation to solve for Δy:
Δy = (v₀²) / (2g)
Plug in the given values:
Δy = (13²) / (2 * 9.8)
Δy = 169 / 19.6
Δy ≈ 8.63 m
Therefore, the skier reaches a height of approximately 8.63 meters above the top of the ramp.
To determine how high the skier gets above the top of the ramp, we can use the principles of projectile motion. The vertical motion can be analyzed separately from the horizontal motion.
First, let's assume that the initial height of the skier from the ground is zero.
We can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (which is zero at the top of the flight)
vi = initial velocity (13 m/s upwards)
a = acceleration due to gravity (-9.8 m/s^2, as its direction is downwards)
d = displacement (the height we want to find)
Rearranging the equation, we have:
0^2 = (13 m/s)^2 + 2(-9.8 m/s^2)d
Simplifying, we get:
0 = 169 m^2/s^2 - 19.6 m/s^2 * d
Rearranging again, we have:
19.6 m/s^2 * d = 169 m^2/s^2
Solving for d (height), we get:
d = 169 m^2/s^2 / 19.6 m/s^2
d ≈ 8.622 m
Therefore, the skier reaches a height of approximately 8.622 meters above the top of the ramp.
depends on the angle of the ramp at takeoff.
H = v^2/(2g) sin^2θ
all we know so far is that v cosθ = 13