Question

What is the maximum height achieved if a 0.600 kg mass is thrown straight upward with an initial speed of 75.0 m·sâ1? ignore the effect of air resistance?

Answer 1

We can solve the problem by using the law of conservation of energy:

- at the beginning, all mechanical energy of the object is just kinetic energy: $K=\frac{1}{2}mv^2$, where m is the mass and v is the velocity

- at the point of maximum height, all mechanical energy of the object is just gravitational potential energy: $U=mgh$, where h is the maximum height

Therefore, the conservation of energy becomes:

$\frac{1}{2}mv^2 = mgh$

Re-arranging, we find the maximum height:

$h=\frac{v^2}{2g} = \frac{(75.0 m/s)^2}{2(9.8 m/s^2)}=287.0 m$