Question

A daredevil college professor wants to jump across a canyon of depth 100m and width 50m on his motorcycle. He uses a ramp inclined at 60 degrees to cross the canyon.The other side of the canyon is 10m lower than his side. Find the Minimum velocity, vo, that his motorcycle should have to make it safely across the canyon. If he jumps with a velocity of vo/3 find out what (x,y) position he would crash

Answer 1

Let us consider the initial velocity is V_o
now horizontal component of the velocity (V_x) = V_oCos = V_oCos60
Vertical component of the velocity (V_y) =VSin = V_oSin60
First considering the horizontal motion
Distance covered in the horizontal direction (R) =50 m
Time taken to cover this distance (t) =R/V_X = (50/V_oCos60) = 100/V_o -----------(1)
now the vertical motion
Distance covered in vertical direction = 10 m
S = ut +(1/2)at²
10 = V_y*t +(1/2)*9.81*t²
-10 = V_oSin60*(100/V_o) -(1/2)*9.81*(100/V_o)²
= 86.602 - 49050/V_o²
49050/V_o² = 86.602 -10 = 76.602
V_o² = 49050/76.602
V_o = 25.304 m/s
hence the initial velocity should be 25.304 m/s to reach at the other point of the canyon.
(ii)Now if the velocity would be = V_o/3 = 25.304 /3 = 8.435 m/s
Considering vertical motion
-100 = 8.435Sin60*t -(1/2)*9.81*t²
-100 = 7.305t - 4.905t²
4.905t² -7.305t -100 = 0
on solving it we get
t = 5.321 s
Horizontal distance covered in this time = 8.435Cos60*5.321 = 22.44 m
hence the x = 22.44 m and y = 100 m