Sam (85 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis. The skis have a thrust of 160 N. He keeps his skis tilted at 10° after becoming airborne. How far does Sam land from the base of the cliff?
2007-10-07 04:47:52 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Acceleration along the ramp
a = 160/m - gsin(theta)
Velocity at the end of the ramp
v0 = sqrt(2as) = sqrt(2a*50/sin(theta))
Vertical velocity
v0y = v0sin(theta)
Hor. velocity
v0x = v0cos(theta)
In-flight vertical acceleration
ay = 160sin(theta)/m - g
Peak height
ymax = 50 + v0y^2/(2*ay)
Flight time to peak height
t1 = v0y/ay
Flight time from peak height to landing
t2 = sqrt(2ymax/ay)
Total flight time
t = t1 + t2
In-flight horizontal acceleration
ax = 160cos(theta)/m
Hor. distance from the base
x = v0x*t + ax*t^2/2
2007-10-08 04:13:01 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
Samland
2016-12-16 15:54:05 · answer #2 · answered by Anonymous · 0⤊ 0⤋