Julie starts at the top of the 60 degree slope shown in the figure(Figure 1) . A

Question

Julie starts at the top of the 60 degree slope shown in the figure(Figure 1) . At the bottom, a circular arc carries her through a 90 degree turn, and she then launches off a 3m high ramp. How far horizontally is her touchdown point from the end of the ramp?

Julie starts at the top of the 60 degree slope shown in the figure(Figure 1) . At the bottom, a circular arc carries her through a 90 degree turn, and she then launches off a 3m high ramp. How far horizontally is her touchdown point from the end of the ramp?

Explanation / Answer

ulie's only energy is coming from her gravitational potential energy. Since she starts from 25m high and launches from 3m high, the total vertical falling distance is 22m, (25-m 3m =22m)

Since she is on a frictionless slope, her velocity leaving ramp will be equal to herfreefall velocity of falling 22m. Plus, the exit ramp is set at 30 deg, (180deg -60deg - 90deg= 30 deg). Her freefall velocity is just being redirected on the circular arc, so her exit velocity off the ramp is the same velocity as her freefall velocity after 22m:

Vy = vertical velocity
Vy = gt
y = 1/2gt
Vy = 20.77 m/s = launch speed from ramp at 30 deg

using equations:
x = Vox*t
Vox = velocity in x direction
t = time

y = Voy *t - 1/2gt^2
Voy = velocity in y direction
g = 9.8m/s^2
substitute t = x / Vox
y = xVoy / Vox - 1/2g*x^2 / (Vox)^2

Range =R = Vo^2sin2{theta} / g

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