Projectile Motion

In projectile motion the horizontal and vertical motions are independent, with constant horizontal velocity and vertical motion governed by gravity. The maximum height is reached when the vertical component of velocity momentarily becomes zero. The vertical launch component is $u_y=u\sin \theta =20\sin 30^{\circ }=20\times 0.5=10\text{ m/s}$. Using $v_y^2=u_y^2-2gH$ with $v_y=0$ at the peak gives $H=\frac{u_y^2}{2g}=\frac{10^2}{2\times 10}=\frac{100}{20}=5\text{ m}$. The option 10 m is wrong because it forgets to square and halve correctly, effectively using $u_y^2/(2g)$ with $u_y=20$ wrongly. The option 20 m ignores the angle and uses the full launch speed vertically. The option 2.5 m wrongly divides by an extra factor of two. This follows the NCERT projectile formula $H=\frac{u^2{\sin }^2\theta }{2g}$. It is worth emphasising that the horizontal component $u\cos \theta$ continues unchanged throughout the flight and plays no part in determining the peak height, which is governed entirely by the vertical launch component and gravity. As a check, only the vertical velocity component governs height, so a low launch angle of 30 degrees sensibly yields a modest peak well below the height a fully vertical launch of the same speed would reach.