Height of the waterfall is 0.449 m
its horizontal distance will be 2.1 m
now let say his speed is v with which he jumped out so here the two components of his velocity will be
[tex]v_x = vcos35.2 = 0.817 v[/tex]
[tex]v_y = vsin35.2 = 0.576 v[/tex]
here the acceleration due to gravity is 9.81 m/s^2 downwards
now we can find the time to reach the other end by y direction displacement equation
[tex]\Delta y = v_y * t + \frac{1}{2} at^2[/tex]
[tex]-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2[/tex]
also from x direction we can say
[tex]\Delta x = v_x * t[/tex]
[tex]2.1 = 0.817 v* t[/tex]
now we have
[tex] v* t = 2.57[/tex]
we will plug in this value into first equation
[tex]- 0.449 = 0.576 * 2.57 - 4.905 * t^2[/tex]
[tex]1.93 = 4.905 * t^2[/tex]
[tex]t = 0.63 s[/tex]
now as we know that
[tex]v* t = 2.57 [/tex]
t = 0.63 s
[tex]v = \frac{2.57}{0.63}[/tex]
[tex]v = 4.1 m/s[/tex]
so his minimum speed of jump is 4.1 m/s
