Sam (75 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis.?

Question

Sam (75 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis. The skis have a thrust of 240 N. He keeps his skis tilted at 10° after becoming airborne, as shown in Figure P6.43. How far does Sam land from the base of the cliff?

Answer 1

Fig P6.43 OK I see...

1. What is his speed at the top. The horizontal component is the one I'm after.
2. Assume that the top of cliff and the top of the slope is same.

1. The kinetic energy at the top
Ke=0.5mV^2
V= sqrt(2Ke/m)
Ke=work done = F s
where s=50/sin(10)=288 m
so Ke= 240 x 288=69105 Joules

Now Vh=V cos(10)
Vh= sqrt( 2Ke/m) cos(10)

But this is not all the Vh additional V'h is supplied by the jet-powered skis after he is airborne.

V'h= at cos(10) where a=F/m
and we should not forget the acceleration component directed upwards either.

a'=(F/m)sin(10)

Now we need the the time

t=sqrt(2h/(g-a') where
h=50m
g=9.81m/s^2

Now finally

S=(Vh + V'h)t

S=(Vh +at cos(10) t)
S=Vh +(F/m) t^2 cos(10)