Question

an object thrown up with an instant velocity 40m/s from grounds. assume the energy is conservative. find the maximum height

Answer 1

$E_k+Ep= const$

$mgh+\frac{mV^2}{2}= const$

$\text{where }E_k,E_p \text{kinetic and potential energy of the object}$

$g=9.8m/s^2$

$\text{during the throw:}$

$h_0 = 0;V= V_0=40m/s$

$E_k+Ep= mgh_0+\frac{mV_0^2}{2}=\frac{mV_0^2}{2}$

$\text{when the object reaches its maximum height:}$

$V= 0;E_k+Ep= mgh_{max}$

$\text{compare both parts}$

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

$h_{max}=\frac{V_0^2}{2g}=\frac{40^2}{2*9.8}\approx16.66$

Answer:$h_{max}=16.66m$