Skiing off of Roof Physics Question: Looking for time in the air, whether the skier clears a road, and final velocity on landing

Question

Mac is attempting to film himself ski down the roof for inclusion in the latest installment of his line of videos featuring ill-conceived stunts. The snowy roof is inclined at 30 degrees below the horizontal, with the lower edge being 10 ft above the ground. Mac accelerates from rest at half the free fall rate along the roof, skiing 8 ft before becoming airborne. A slight panic sets in as he notices the busy street below, 8 ft away horizontally from the edge of the roof. (a) For how long does Mac's life flash before his eyes? (b) Is this Mac's last stunt? (c) How fast is Mac moving upon landing?

Answer 1

a)There are two major steps for part a)

• a1) Strategy: Since Vo, a and d are given use kinematics equation to find launching velocity V
• a2) Right after launching it is a free fall (acceleration = g = -10 m/s). From V (found in a1) use the vertical component Vy as the initial velocity downwards. You'll probably have Vy=-V*sin(30). Use the y=yo+Vyt+(1/2)at^2 equation to find the time.

 

b)Depends. Use the time (found in a2) and x=vt to calculate the horizontal distance. If X<8ft, Mac is safe.

 

c) You need both components Vfx and Vfy and then Pythagorean theorem to find Vf. By the way, Vfx=Vx (horizontal component of launching velocity), Vfy can be found using Vfy=Vy+at where a=-10m/s^2

 

If you have any questions, please do not hesitate to contact me. Best, Martin.