Question

A body is projected at an angle of ${30}^o$ to the horizontal with kinetic energy $40J$. What will be the kinetic energy at the top-most point?

• A. $20J$
• B. $25J$
• C. $30J$
• D. $40J$

Answer 1

Answer: (C)

$30J$

Let the initial speed of projectile be $u$.
Initial kinetic energy of projectile $K.E=\frac{1}{2}m{u}^2=40J$
Horizontal component of velocity $u_x=u\cos \theta =u\cos {30}^o$
At the highest point, vertical component of velocity vanishes and only the horizontal component of velocity remains.
So, kinetic energy of projectile at the top-most point $K.E^{\prime }=\frac{1}{2}m{u}_x^2$
$⟹K.E^{\prime }=\frac{1}{2}m{u}^2{\cos }^2{30}^o=40\times (0.866{)}^2=30J$