Vo = 10m/s[30o]
Xo = 10*Cos30 = 8.66 m/s
Yo = 10*sin30 = 5 m/s
1. Y = Yo * g*Tr = 0
Tr = -Yo/g = -5/-9.8 = 0.51 s. = Rise time.
h = ho + -(Yo^2)/2g = 19 + -(5^2)/-19.6 = 20.3 m. Above gnd.
h = 0.5g*t^2 = 20.3
4.9t^2 = 20.3
t^2 = 4.14
Tf = 2.03 s. = Fall time.
Dx = Xo*(Tr+Tf) = 8.66 * (0.51+2.03) =
22 m. Which is greater than the required
hor. distance of 14 m.
2. Y^2 = Yo^2 + 2g*h=0 + 19.8*(20.3-16) = 84.28
Y = 9.2 m/s = Ver. component.
V = sqrt(Xo^2+Y^2) m/s
Xo = 8.66 m/s
Y = 9.2 m/s
Solve for V.
A snowboarder is trying to gap a 14m
wide road. He leaves the 19m high cliff at
10m/s at 30 degrees.
Questions:
1. Does he make it?
2.When the snowboarder is 16m above
the road, what is his speed?
1 answer