The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is:

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Accepted Answer

1. **Velocity Components:** A projectile launched with initial speed $u$ at an angle $	heta$ has a horizontal velocity component $u_x = u \cos 	heta$ and a vertical component $u_y = u \sin 	heta$.
2. **Velocity at Maximum Height:** At the maximum height, the vertical component of the velocity becomes zero ($v_y = 0$). Therefore, the speed of the projectile at maximum height is purely its horizontal component, which remains constant throughout the flight (neglecting air resistance) .
   $$ v_{top} = u_x = u \cos 	heta $$
3. **Given Condition:** The speed at maximum height is half of the initial speed.
   $$ v_{top} = \frac{u}{2} $$
4. **Calculation:** Equating the expressions:
   $$ u \cos 	heta = \frac{u}{2} $$
   $$ \cos 	heta = \frac{1}{2} $$
   $$ 	heta = \cos^{-1}(0.5) = 60^{\circ} $$

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