Respuesta :
initial speed of the stuntman is given as
[tex]v = 28 m/s[/tex]
angle of inclination is given as
[tex]\theta = 15 degree[/tex]
now the components of the velocity is given as
[tex]v_x = 28 cos15 = 27.04 m/s[/tex]
[tex]v_y = 28 sin15 = 7.25 m/s[/tex]
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
[tex]\delta y = -25 m[/tex]
[tex]\delta y = v_y * t + \frac{1}{2} at^2[/tex]
[tex]-25 = 7.25 * t - \frac{1}{2}*9.8* t^2[/tex]
by solving above equation we have
[tex]t = 3.12 s[/tex]
Now in the above interval of time the horizontal distance moved by it is given by
[tex]d_x = v_x * t[/tex]
[tex]d_x = 27.04 * 3.12 = 84.4 m[/tex]
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.
