Answer:
(a) [tex]v_f= 26.7095 m/s[/tex]
(b) 5.615979
(c) [tex]v_f= 26.82624 m/s[/tex]
(d) t=5.114872571 seconds
Explanation:
(a)
Acceleration, a along the slope is given by
[tex]a=gsin\theta[/tex] where g is acceleration due to gravity and [tex]\theta[/tex] is the angle of inclination
Also, the velocity after moving a distance s is given by
[tex]v_f^{2}=v_i^{2}+2as[/tex] where [tex]v_f[/tex] and [tex]v_i[/tex] are final and initial velocities, s is distance covered
Substituting [tex]a=gsin\theta[/tex] into the above we obtain
[tex]v_f^{2}=v_i^{2}+2sgsin\theta[/tex]
Making [tex]v_f[/tex] the subject we obtain
[tex]v_f=\sqrt{v_i^{2}+22sgsin\theta}[/tex] and substituting [tex]9.81 m/s^{2}[/tex] for g, [tex] 29^{0}[/tex] for [tex]\theta[/tex] and 75 m for s while initial velocity is zero
[tex]v_f=\sqrt {2sgsin\theta}=\sqrt {2*75*9.81sin29}[/tex]
[tex]v_f= 26.7095 m/s[/tex]
(b)
Time to reach the bottom is given by
[tex]v_f=v_i+at=v_i+gtsin\theta[/tex] and making t the subject of the formula
[tex]t=\frac {v_f-v_i}{gsin\theta}[/tex] but since initial velocity is zero hence
[tex]t=\frac {v_f}{gsin\theta}=\frac {26.696}{9.81sin29^{o}}= 5.615979 seconds[/tex]
(c)
When the initial velocity is 2.5 m/s then
[tex]v_f=\sqrt {V_i^{2}+2sgsin\theta}[/tex]
[tex]v_f=\sqrt {2.5^{2}+(2*9.81*75sin29^{o}}[/tex]
[tex]v_f= 26.82624 m/s[/tex]
(d)
Time required to reach the bottom is given by
[tex]t=\frac {v_f-v_i}{a}=\frac {v_f-V_i}{gsin\theta}[/tex] and substituting the above figures we get
[tex]t=\frac {26.82624-2.5}{9.81sin29^{o}}[/tex]
t=5.114872571 seconds
