Question

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?

Answer 1

(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

Answer 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