A 2.0-kg projectile is fired with initialvelocity components \(\displaystyle{v}_{{{0}{x}}}={30}\) m/s and \(\displaystyle{v}_{{{0}{y}}}={40}\) m/s from a point on the earth's surface. Neglect any effects dueto air resistance.

\(\displaystyle{v}^{{2}}={\left({v}_{{{0}{x}}}\right)}^{{2}}+{\left({v}_{{{0}{x}}}\right)}^{{2}}\)

\(\displaystyle{v}={50.0}\ \frac{{m}}{{s}}\)

Angle for projectile (q)

\(\displaystyle{V}_{{{0}{x}}}={v}{\cos{{\left({q}\right)}}}\)

\(\displaystyle{\left({q}\right)}={53.13}^{\circ}\)

The height attained by the object \(\displaystyle{\left({h}\right)}={\frac{{{\left[{V}\cdot{\sin{{\left({q}\right)}}}\right]}^{{2}}}}{{{2}{g}}}}\)

\(\displaystyle={81.63}\ {m}\)

The kinetic energy of the projectile when it reaches thehighest point in its trajectory is equal to the potential energy

\(\displaystyle{K}.{E}={m}{g}{h}\)

\(\displaystyle={1600}\ {J}\)

The work-done infiring the projectile \(=1600 J\)