Skier leaves a jump with velocity of 22.4m/s at angle of 30. Landing point is 78.6m lower than the starting pt

Question

Find
a) Total distance skier jumped
b) velocity when she landed

full solution please

Answer 1

the first step is to separate the horizontal motion, which has no acceleration when we ignore air resistance, and vertical motion, which has an acceleration of -g. In metric units at 3 sig figures, 9.81 m/s^2.

Using a bit of trig, the horizontal velocity is constant at
vx=22.4*cos(30)

Which means that horizontal position is
x(t)=vx*t, where t is the time since take-off

and the vertical velocity, assuming that the jump is angled upward,
vy(t)=22.4*sin(30)-g*t

Note that at t=0 gravity has had no effect. Also note that the velocity will steadily decrease to zero and then become negative.
since the landing point is 76.8 m below that starting point,
y(t)=
78.6+22.4*sin(30)*t-.5*g*t^2


Okay, now working from the knowledge that when the skier lands y=0, solve for t. There will be two roots, one positive, that's the one you want, and one negative. The negative root can be explained. Send me an email if you care to have the explanation.

Once you know t, you can find vy(t)
Her velocity is the resultant vector with magnitude
Sqrt(vy(t)^2+vx^2)

j

Answer 2

break the velocity into x and y components

Vx = cos(30) 22.4 = 19.4 m/s
Vy = sin(30) 22.4 = 11.2 m/s

a) Vf = at + Vi

we have inital veloicty and acceleration but not final velocity when the skier land vertically

Vf^2 = 2ad + Vi^2
Vf^2 = 2(-9.8)(0 - 78.6) + 11.2^2
Vf^2 = 1666
Vf = -40.82 m/s (the negative means the motion is downward)

-40.82 = (-9.8)t + 11.2
-52.02 = -9.8t
t = 5.3s

So the total time is 5s.

X = vt
X = (19.4) (5.3)
X = 102.82m

b) i just found it in part A but that speed was a vertical speed. If the question ask for the speed at an angle than just use pythagarean theorem