A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy ​

Question

A ball having kinetic energy KE, is projected at an angle of \(60^{\circ}\) from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

(1) \(\frac{(\mathrm{KE})}{8}\)

(2) \(\frac{(\mathrm{KE})}{4}\)

(3) \( \frac{(\mathrm{KE})}{16}\)

(4) \(\frac{(\mathrm{KE})}{2}\)

Answer 1

Correct option is : (2) \(\frac{(\mathrm{KE})}{4}\) 

Initial K.E,

\( K.E. =\frac{1}{2} \mathrm{mu}^{2}\) 

Speed at heighest point

\( \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}\) 

\( \therefore \ \mathrm{KE}_{2}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^{2}\) 

\( =\frac{1}{4} \times \frac{1}{2} \ \mathrm{mu}^{2}\) 

\( =\frac{\mathrm{KE}}{4}\)