How far does Sam land from the base of the cliff?

[xstetsonx]

Sam (60 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis. The skis have a thrust of 200 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?

a=F/m


I used 50/sin10 to get the length of the slope-287.94m. Then i calculated the f on the slope. 200N-mgsin10. then i use that to divided by the mass which gives 1.63 for a. after that i go the final velocity before it jump off. then i have no idea wat to do next

 

[ideasrule]

How long does he stay airborne for? How far, horizontally, does he travel in that time? If you don't know, look at your equations for kinematics and projectile motion.

 

[xstetsonx]

i know i need to get the time from the y-axis then calculate the distance on x axis. but i am not sure wat do i put for Vi in the y axis

 

[ideasrule]

I thought you already calculated the speed that Sam leaves the cliff at. Just do Vsin10 to find the vertical component of his speed and that's your Vi.

 

[xstetsonx]

I thought you already calculated the speed that Sam leaves the cliff at. Just do Vsin10 to find the vertical component of his speed and that's your Vi.

so t=-50=(30.6528sin10)t+.5(-9.8)t^2
t=3.78
then for the x axis
(30.6528cos10)3.78+.5(200/60)(3.78^2)
the acceleration on x-axis is just 200/60 right