g h = (1/2) s^2
h is meters of takeoff point below top
g is about 9.81
s is speed magnitude in m/s
solve for s = sqrt (2 g h)
now
u is horizontal speed
vi is vertical speed up at launch off ramp
V is speed vector = u i + vi j
u = s cos 21
vi = s sin 21
now we have a falling problem
u = s cos 21 forever
v = vi - g t
h = Hi + vi t - 4.9 t^2
if Hi - h = 33
then
0 = 33 + vi t - 4.9 t^2
or
t^2 - vi t - 33 = 0
solve quadratic for t
then
horizontal distance = u t
A ski jump's velocity can be calculated (assume no friction).
The tower is 60.0m high and the skier is at the bottom, who will leave the jump under a 21 degree angleto the horizontal, and land 33m below the starting point. The skier has the exact same velocity at the takeoff as they do at the bottom of the straight ramp.
I need to do these questions that go with the exersise:
1. What is the takeoff velocity of the ski jumper at the bottom of the ramp? (hint: magnitude and direction)
2. How far will the ski jumper jump with the information given and simplifying assumptions made?
1 answer
1 answer