an object thrown up with a instant velocity 40m/s from grounds. assume the energy is conservative. find a) maximum height
b)when where does PE = KE if mass m=2kg
an object is thrown up with an instant velocity from grounds that is to say 40m/s from grounds
"v_o" = 40m/s. if we assume energy is conservative when an object reaches maximum height
the kinetic energy is equal to 0 so "H_m = \\frac{v_o^2}{2g} = \\frac{1600}{2*10} = 80m"
the maximum height is 80m.
2) the second question is not completely given because PE = KE is points and the picture must be given
or PE=KE in question is not fully described
Comments
Leave a comment