question_answer

A projectile is fired with kinetic energy 1 kJ. If the range is maximum, what is its K.E. at the highest point?

Answer:

Here, \[\frac{1}{2}m{{\upsilon }^{2}}=1kJ=1000J\]. For maximum range, \[\theta ={{45}^{\text{o}}}\] and velocity of projectile at the highest point = \[\upsilon \cos {{45}^{\text{o}}}\]. At the highest point, \[KE=\frac{1}{2}m{{\left( \upsilon \cos {{45}^{\text{o}}} \right)}^{2}}=\frac{1}{2}m\frac{{{\upsilon }^{2}}}{2}=\frac{1000}{2}=500J\]