A skier leaves a ski jump with a horizontal velocity of 20.0m/s (and no vertical velocity component ). If the height is 10.0 m,
(a) How long will it take for the skier to land?
(b) What is the magnitude of the horizontal component of the skier's displacement "D" furing the jump?
(c) What is the magnitude of the veritcal component of velocity "Vy" the instant before it she land?
(d) What is the magnitude of the horizontal component of velocity "Vx" the instant before it she land?
(e) What is the magnitude of her velocity "v" the instant before she lands?
(a) h = -4.9 t^2 + 10
... set h=0 and solve for t
(b) horizontal velocity remains constant
... displacement = 20.0 * t
(c) Vy = g * t
(d) see (b)
(e) v^2 = (Vx)^2 + (Vy)^2
Given:
Y = 0 = H
Y_0 = 10.0 = H_0
(a) y = y_0 + v_0*t - 1/2 *gt^2
0 = 10.0 + (0)t - 1/2 *9.8t^2
t = rt(10.0/4.9)
t = 1.43 s
(b) d = vt
D = (20.0)(1.43)
D = 28.6 m
(c) Vy = gt
Vy = (9.8)(1.43)
Vy = 14.0 m/s
(d) Vx = V cos(theta)
Vx = (20.0) cos(0)
Vx = 20.0 m/s
(e) V = rt(V^2x + V^2y)
V = rt(20.0^2 + 14.0^2)
V = 24.4 m/s
To solve these questions, we can use the basic equations of motion. We'll start by finding the time it takes for the skier to land.
(a) To find the time it takes for the skier to land, we can use the equation:
h = (1/2) * g * t^2
where:
h is the height (10.0 m),
g is the acceleration due to gravity (9.8 m/s^2),
and t is the time.
Rearranging the equation to solve for t, we get:
t = sqrt((2 * h) / g)
Substituting the given values, we have:
t = sqrt((2 * 10.0) / 9.8)
t = sqrt(20 / 9.8)
t ≈ sqrt(2.0408)
t ≈ 1.43 s
Therefore, it will take approximately 1.43 seconds for the skier to land.
(b) The magnitude of the horizontal component of displacement "D" can be found using the equation:
D = v * t
where:
v is the horizontal velocity (20.0 m/s), and
t is the time (1.43 s).
Substituting the given values, we have:
D = 20.0 * 1.43
D ≈ 28.6 m
Therefore, the magnitude of the horizontal component of the skier's displacement "D" during the jump is approximately 28.6 meters.
(c) The magnitude of the vertical component of velocity "Vy" at the instant before she lands will be equal to the final vertical velocity. Since the skier starts with no vertical velocity and falls down due to gravity, the vertical component of velocity "Vy" will be equal to the acceleration due to gravity.
Thus, Vy = g = 9.8 m/s^2.
Therefore, the magnitude of the vertical component of velocity "Vy" at the instant before landing is 9.8 m/s^2.
(d) The horizontal component of velocity "Vx" remains constant throughout the motion. Therefore, the magnitude of the horizontal component of velocity "Vx" at the instant before landing will be the same as the initial horizontal velocity of the skier.
Hence, Vx = 20.0 m/s.
Therefore, the magnitude of the horizontal component of velocity "Vx" at the instant before landing is 20.0 m/s.
(e) The magnitude of the velocity "v" at the instant before landing can be found using the Pythagorean theorem:
v = sqrt(Vx^2 + Vy^2)
Substituting the given values, we have:
v = sqrt((20.0^2) + (9.8^2))
v = sqrt(400 + 96.04)
v = sqrt(496.04)
v ≈ 22.27 m/s
Therefore, the magnitude of the velocity "v" at the instant before landing is approximately 22.27 m/s.
To solve these problems, we can use the principles of kinematics. In this case, we are given the horizontal velocity, the height of the ski jump, and we assume that air resistance is negligible.
(a) To calculate the time it takes for the skier to land, we can use the equation of motion for vertical motion:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Plugging in the values, we have:
10.0 m = (1/2)(9.8 m/s^2)t^2
Solving for t, we get:
t^2 = (2 * 10.0 m) / 9.8 m/s^2
t^2 = 2.04 s^2
Taking the square root of both sides:
t ≈ 1.43 s
Therefore, the skier will take approximately 1.43 seconds to land.
(b) The magnitude of the horizontal component of displacement D is equal to the horizontal velocity multiplied by the time. Given that the horizontal velocity is 20.0 m/s, and the time is 1.43 s, we have:
D = (20.0 m/s) * (1.43 s)
D = 28.6 m
Therefore, the magnitude of the horizontal component of the skier's displacement during the jump is 28.6 meters.
(c) At the instant just before the skier lands, the vertical component of velocity Vy is equal to zero, as the skier has reached the maximum height of the jump and is about to descend. Therefore, the magnitude of Vy is 0 m/s.
(d) The horizontal component of velocity Vx remains constant throughout the motion. Therefore, the magnitude of Vx just before landing is still 20.0 m/s.
(e) To find the magnitude of the skier's velocity v just before landing, we can use the Pythagorean theorem. Given that Vx = 20.0 m/s and Vy = 0 m/s, we have:
v = sqrt(Vx^2 + Vy^2)
v = sqrt((20.0 m/s)^2 + (0 m/s)^2)
v = sqrt(400 m^2/s^2)
v ≈ 20.0 m/s
Therefore, the magnitude of the skier's velocity just before landing is approximately 20.0 m/s.