A ski jumper leaves a slope at an angle of 20.0 degrees above the horizontal direction. She lands 3.50

Question

seconds later at a point 20.0 meters below her take-off point.
(a) What was her initial speed?
(b) How far does she travel horizontally?

Answer 1

h(t) = vt - 1/2 g t^2
3.5v - 4.9*3.5^2 = -20
v = 11.436

That v is the vertical speed at takeoff. So, the entire initial speed is

11.436/sin20° = 33.437

Use that speed and

d = vt to get the horizontal distance traveled.

d = 33.437 * 3.5 = 117.028

Answer 2

oops. the horizontal speed is 33.437 cos20°

make that fix...