Answer to Question #154842 in Mechanics | Relativity for mule

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 Porential energy= Kinetic Energy if mass m=2kg

Answer 1

$E_p =mgh\text { where }g=9.8m/s^2$

$E_k=\frac{mV^2}{2}$

$\text{if }h =0 \text{ then }V_0=40\text{ by problem statement}:$

$E_p+E_k= mg*0+\frac{mV_0^2}{2}=\frac{mV_0^2}{2}$

$\frac{mV_0^2}{2} - \text{this value is constant for } E_p+E_k$

$a) \text{for }h_{max}\ V=0:$

$E_p+E_k= mgh_{max}+\frac{m*0}{2}= mgh_{max}$

$mgh_{max}=\frac{mV_0^2}{2}$

$h_{max}=\frac{V_0^2}{2g}=\frac{40^2}{2*9.8}\approx81.6$

$b)E_p=E_k$

$2E_p = \frac{mV_0^2}{2}$

$E_p = \frac{mV_0^2}{4}= \frac{2*40^2}{4}=800$

$E_k=800$

Answer: $a)h_{max}\approx81.6m\ b)\text{if }E_p=E_k \text{ then}E_p=E_k=800J$