# A 2.0-kg projectile is fired with initial velocity components v_{0x}=30 m/s and v_{0y}=40 m/s from a point on the earth's surface. Neglect any effects

A 2.0-kg projectile is fired with initial velocity 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 due to air resistance. What is the kinetic energy of the projectile when it reaches the highest point in its trajectory? How much work was done in firing the projectile?

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Arnold Odonnell

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 $$=1600 J$$