initial speed of the stuntman is given as
angle of inclination is given as
now the components of the velocity is given as
here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.
So the displacement in vertical direction is given as
by solving above equation we have
Now in the above interval of time the horizontal distance moved by it is given by
since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.
Try to draw Humpty Dumty cracked on the floor with men all around him
Answer:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN
With another person on student A’s side. I’m sorry if I wasn’t able to help. Im trying my best from the information given. :)
