Question

A skater rides off a drop of 35 degrees at 2.0m/s. When do they land on the decline?

Answer 1

The slope is tan(35) right?

Answer 2

"when do they land on the decline" is that a 'time' question or a 'where on the incline do they land' question?

Answer 3

@doc.brown

Answer 4

y-direction:
s=s+vt+(1/2)at^2
y = 0 + 0 + (1/2)(-9.81)t^2
y=4.9t^2 [eq 1]

x-direction
s=s+vt+(1/2)at^2
x = 0 + 2(t) + (1/2) (0) t^2
x = 2t [eq 2]

use angles given & combine eq 1 and eq 2
tan(35) = y/x
tan(35) = ((-9.8/2) * t^2 ) / (2*t)
t = 0.285 sec

get values for x and y using eq 1 and 2
y=4.9t^2 [eq 1]
y=4.9(0.285)^2
y=0.39m
x = 2t [eq 2]
x = 2(0.285)
x = 0.57m

find distance using pathagorean's therome
x^2 + y^2 = d^2
0.57^2 + 0.39^2 = d^2
d = 0.69
d = 0.7 meter