1At the instant of leaving the ramp the skier has a speed of

1)At the instant of leaving the ramp, the skier has a speed of 20 m/s at an angle ?A = 30° with the horizontal. In addition to gravity, the wind gust exerts a horizontal acceleration of 2t m/s2 to the left on the skier. Determine the flight time for the skier to strikes the ground. [Hint: You may need to find the root of a higher-order polynomial].

Solution

verticla vel = 20Sin(30) = 10m/s

Horizontal vel. 20Cos(30) = 17.32 m/s

let t be the time of flight

Horizontal dispalcement x is given by

2ax= u²

intial horizontal vel u = 17.32 m/s

horizontal acceleration = -2t m/s ( wind gust)

x = 17.32²/4t = 75/t

the ram p slope is 3/4

y/x = 3/4

3x = 4y --------(1)

during this time the skier has gone up and came back tot he starting position and reacheed B under gravity

intial vertical vel = 10 m/s

time of vertical raise is givne by

v= gt\'

t\' = 10/10 = 1 ( we take g = 10m/s for simplicity)

height of ascent s = 10^2/2*10 = 5 m

The skier has desenced under gravity for (t-t\' =t-1) seconds

height descended = gt²/2 = 10*(t-1)²/2 = 5(t-1)²

The totals hieght descended = y+8+5 = y+13

y+13 = 5(t-1)²

we have y = 3x/4 from (1) eliminate y and x from above we have

(3/4)(75/t) +13 = 5(t-1)²

20t³ - 40t² -32t -225 = 0

you get the value of t- time of flight by solving the above equation.