vx = horizontal speed = 25 m/s
Θ = inlcination of the plane = 33°
g= gravity = 9.8 m/s^2
vy = vertical speed = 0 m/s
Where:
x = horizontal displacement before landing on the inclined plane
y= vertical displacement before landing on the inclined plane
d= distance covered by the ski in the air
y= vy t + 1/2 g t^2
y= 0 + 1/2 g t^2
y= 1/2 g t^2
Where t is time spent by the ski jumper in the air
Speed = distance / time
vx = x / t
x = vx t
Apply tangent function
Tan Θ = opposite side / adjacent side
Tan Θ = y /x
Tan Θ = (1/2 g t^2) / (vxt)
Tan Θ = gt / 2vx
Solve for t
t = 2vx tan(33) / 9.8
t= 3.3 seconds (a)
Part B. where does he land?
Apply the Pythagorean theorem:
hypotenuse^2 = side1^2 + side2 ^2
d^2 = x^2 + y^2
d^2 = (1/2 g t ^2)^2 + (vx t )^2
d^2 = (1/2 * 9.8 * 3.3^2 ) + (25*3.3)^2
d= √ (1/2 * 9.8 * 3.3^2 ) + (25*3.3)^2
d= 98 meters
