A ball is thrown at an angle θ with the horizontal. Its initial kinetic energy is 100 J and it becomes 30 J at the highest point. The angle of projection is

The correct option is cos⁻¹(√(3/10)).

Let the mass of the ball be $m$ and its initial velocity of projection be $v$ at an angle $\theta$ with the horizontal.

The initial kinetic energy ($K_i$) of the ball is given by:
$K_i=\frac{1}{2}m{v}^2$
According to the problem, the initial kinetic energy is 100 J. Therefore:
$\frac{1}{2}m{v}^2=100\text{J}$ (Equation 1)

At the highest point of its projectile trajectory, the vertical component of the velocity becomes zero ($v_y=0$). Only the horizontal component of the velocity remains, which is constant throughout the motion:
$v_h=v\cos \theta$

Therefore, the kinetic energy at the highest point ($K_h$) is:
$K_h=\frac{1}{2}m{v}_h^2=\frac{1}{2}m{\left(v\cos \theta \right)}^2$
$K_h=\left(\frac{1}{2}m{v}^2\right){\cos }^2\theta$

Given that the kinetic energy at the highest point is 30 J, we can substitute $K_h=30\text{J}$ and the value of $\frac{1}{2}m{v}^2$ from Equation 1:
$30=100{\cos }^2\theta$

Now, we solve for $\theta$:
${\cos }^2\theta =\frac{30}{100}=\frac{3}{10}$
Taking the square root on both sides:
$\cos \theta =\sqrt{\frac{3}{10}}$
Thus, the angle of projection is:
$\theta ={\cos }^{-1}\left(\sqrt{\frac{3}{10}}\right)$