Question

k is the kinetic energy of a projectile fired at an angle 45°. then what is the
kinetic energy at the highest point?​

Answer 1

Explanation:

hey mate,

$k \: = \frac{1}{2} {mu}^{2}$

at highest point velocity become

$\frac{u}{ \sqrt{2} }$

${k}^{.} = \frac{1}{2} m {( \frac{u}{ \sqrt{2} } )}^{2}$

${k}^{.} = \frac{k}{2}$

hope this would help you!!