st height, the vertical velocity is zero. The particle is still moving horizontally, but momentarily it is not moving up or down. Use vertical SUVAT only: $$ v_y^2 = u_y^2 + 2as $$ At the top, $v_y=0$, $a=-g$, and $s=H$, where $H$ is the height gained above the launch point. A particle is projected from level ground with speed 22 m s⁻¹ at an angle of $50^\circ$ to the horizontal. Find its greatest height above the ground. 1. Find the initial vertical velocity: $$ u_y = 22\sin 50^\circ = 16.85 $$ 2. At the highest point, the vertical velocity is zero: $$ 0 = 16.85^2 - 2(9.8)H $$ 3. Solve for $H$: $$ H = \frac{16.85^2}{19.6} = 14.5 $$ 4. The greatest height is 14.5 m. At maximum height, only the vertical velocity is zero. The horizontal velocity is not zero. ## Time of flight and horizontal range The range is the horizontal distance travelled before the projectile lands. The range of a projectile is the horizontal distance from its launch point to its landing point. If the particle lands at the same height it was launched from, set $y=0$ and solve for the non-zero value of $t$. Then use: $$ x = u_x t $$ A ball is projected from the ground with speed 22 m s⁻¹ at an angle of $50^\circ$. Find how far away it lands. 1. Use the same components: $$ u_x = 22\cos 50^\circ = 14.14 $$ $$ u_y = 22\sin 50^\circ = 16.85 $$ 2. Since it lands back on the ground, vertical displacement is zero: $$ 0 = 16.85t - 4.9t^2 $$ 3. Factorise to find the non-zero time: $$ t(16.85 - 4.9t)=0 $$ $$ t = \frac{16.85}{4.9} = 3.44 $$ 4. Use horizontal motion to find the range: $$ x = 14.14(3.44) = 48.6 $$ 5. The horizontal distance is 48.6 m. The shortcut $R=\frac{U^2\sin 2\theta}{g}$ only works when the projectile lands at the same vertical height as it was launched. ## Starting above the ground If the projectile starts at height $h$ above the ground, use: $$ \text{height above ground} = h + u_y t - \frac{1}{2}gt^2 $$ To find when it hits the ground, set the height equal to zero. A particle is projected from a point 12 m above horizontal ground with velocity $5\mathbf{i}+8\mathbf{j}$ m s⁻¹. Find its greatest height above the ground and the horizontal distance travelled before it lands. 1. Read off the initial components: $$ u_x = 5,\qquad u_y = 8 $$ 2. Find the extra height gained above the launch point: $$ H = \frac{8^2}{2(9.8)} = 3.27 $$ 3. Add the starting height: $$ \text{greatest height above ground} = 12 + 3.27 = 15.3 $$ 4. Set the height above ground equal to zero: $$ 0 = 12 + 8t - 4.9t^2 $$ 5. Rearrange and solve the quadratic, taking the positive root: $$ 4.9t^2 - 8t - 12 = 0 $$ $$ t = 2.58 $$ 6. Use horizontal motion: $$ x = 5(2.58) = 12.9 $$ 7. The particle travels 12.9 m horizontally before landing. ## Finding unknown speed or angle Sometimes you are given the time of flight, greatest height, or a point on the path, and you need to find $U$ or $\theta$. A useful method is to set up equations using the components: $$ u_x = U\cos\theta $$ $$ u_y = U\sin\theta $$ A particle is projected from the ground with speed 30 m s⁻¹ at angle $\theta$ above the horizontal. It lands back on the ground after 4 seconds. Find $\theta$ and the range. 1. Use vertical displacement equal to zero: $$ 0 = (30\sin\theta)(4) - 4.9(4^2) $$ 2. Simplify: $$ 120\sin\theta = 78.4 $$ 3. Find $\theta$: $$ \sin\theta = 0.6533 $$ $$ \theta = 40.8^\circ $$ 4. Use horizontal motion: