Question

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

Answer 1

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