Answer to Question #259983 in Mechanics | Relativity for Jake

Question

A body of mass 1 kg is thrown vertically upward and the body comes back on the earth surface after 8 s. Find the K. E. and potential energy after 2 s. g = 10 ms².

Answer 1

$m = 1kg$

$t_1 = 8s$

$t_2 = 2s$

$g =10\frac{m}{s^2}$

$h(t) = v_0t - \frac{gt^2}{2}$

$h(t_1)=0$

$8v_0-\frac{10*8^2}{2}=0$

$v_0 = 40\frac{m}{s}$

$v(t) = v_0-gt$

$v(t_2) = 40-10*2=20\frac{m}{s}$

$K.E= \frac{mv^2}{2}=\frac{1*20^2}{2}=200J$

$h(t_2)= 40*2-\frac{10*2^2}{2}=60m$

$P.E = mgh = 1*10*60= 600J$

$\text{Answer: }K.E=200J;P.E=600J$