Answer to Question #221149 in Mechanics | Relativity for Neev

Question

a person standing on vertical cliff at height of 180 m above a lake,wantsto jump into the lake but notiecs a rock at the surface level with its furthest edge a distance s from the shore the person realises that with the running start it will be possile to just clear the rock.so the person steps back from the edge a distance d=125/3 and starting freom rest runs at an accelearation that varries in time according to a(x)=2t and then leaves the cliff horizontally.the person just clears the rock.find k if s=k50(take g=10cm/sec(^2) )

Answer 1

$\text{Consider the movement of a person along a cliff:}$

$a(x)=2t$

$a(x) = v'(t)$

$v(t)= t^2+C_1$

$v(0) = 0\implies C_1=0$

$v(t)= t^2$

$s(t) = v'(t)$

$s(t) = \frac{t^3}{3}+C_2$

$s(0) = 0\implies C_2=0$

$s(t) = \frac{t^3}{3}$

$d = \frac{125}{3}$

$s(t_1) =\frac{125}{3}$

$\frac{t_1^3}{3}=\frac{125}{3}$

$t_1=5s;v_x(t_1)=t_1^2=25\frac{m}{s}$

$v_x - \text{speed of a person at the edge of a cliff}$

$\text{Consider a flight from the edge of the cliff}$

$\text{to the surface of the lake:}$

$\text{vertical movement}$

$h = 180m$

$h =\frac{gt_2^2}{2}$

$t_2=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2*180}{10}}=6s$

$\text{horizontal movement}$

$s = v_xt_2$

$s=50k$

$k= \frac{v_xt_2}{50}= \frac{25*6}{50}=3$

$\text{Answer: }k =3$