Answer to Question #102181 in Mechanics | Relativity for bjbnl

Question

A projectile of mass 0.5 kg is shot straight up with an initial speed of 18.8 m/s. How much energy is lost from the system if it reaches a max height of 15.9048 meters?

Answer 1

1). Определим начальную скорость снаряда, при которой он достигнет высоты $h_1=15.9048 m$

To do this, we apply the kinetic energy change theorem

$\frac{mV^2_{01}}{2}=mgh_1$

where do we get

$V_{01}=\sqrt{2gh_1}=\sqrt{2\cdot 9.81\cdot 15.9048}=\sqrt{312.052}=17.665m/c$

2). Then the energy loss is

$\Delta E=\frac{mV^2_{02}}{2}-\frac{mV^2_{01}}{2}=\frac{0.5\cdot 18.8^2}{2}-\frac{0.5\cdot 17.665^2}{2}=88.36-78.01=10.35J$