1. **State the problem:** A projectile is launched from a platform 5 meters above the ground with an initial velocity $v_0 = 40$ m/s at an angle $\theta = 60^\circ$. We need to find the time of flight and the maximum height above the ground. 2. **Identify known values:** - Initial velocity, $v_0 = 40$ m/s - Launch angle, $\theta = 60^\circ$ - Initial height, $h_0 = 5$ m - Acceleration due to gravity, $g = 9.8$ m/s$^2$ 3. **Resolve initial velocity into components:** - Horizontal component: $v_{0x} = v_0 \cos 60^\circ = 40 \times 0.5 = 20$ m/s - Vertical component: $v_{0y} = v_0 \sin 60^\circ = 40 \times \frac{\sqrt{3}}{2} = 20 \sqrt{3}$ m/s 4. **Calculate time of flight:** The projectile rises to a maximum height and then falls to the ground (height 0). Use vertical motion equation: $$ y = h_0 + v_{0y} t - \frac{1}{2} g t^2 $$ At landing, $y=0$: $$ 0 = 5 + 20\sqrt{3} t - 4.9 t^2 $$ Rearranged: $$ 4.9 t^2 - 20\sqrt{3} t - 5 = 0 $$ Use quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=4.9$, $b=-20\sqrt{3}$, $c=-5$: $$ t = \frac{20\sqrt{3} \pm \sqrt{(20\sqrt{3})^2 - 4 \times 4.9 \times (-5)}}{2 \times 4.9} $$ Calculate discriminant: $$ (20\sqrt{3})^2 = 400 \times 3 = 1200 $$ $$ 4 \times 4.9 \times (-5) = -98 $$ So: $$ \sqrt{1200 + 98} = \sqrt{1298} \approx 36.03 $$ Therefore: $$ t = \frac{20\sqrt{3} + 36.03}{9.8} $$ (Since time cannot be negative, take the positive root.) Calculate numerator: $$ 20\sqrt{3} \approx 20 \times 1.732 = 34.64 $$ Sum numerator: $$ 34.64 + 36.03 = 70.67 $$ Calculate time of flight: $$ t = \frac{70.67}{9.8} \approx 7.21 \text{ seconds} $$ 5. **Calculate maximum height:** Maximum height above the ground is initial height plus maximum vertical displacement. Vertical velocity at maximum height is zero: $$ v_y = v_{0y} - g t_{up} = 0 $$ So, $$ t_{up} = \frac{v_{0y}}{g} = \frac{20\sqrt{3}}{9.8} \approx \frac{34.64}{9.8} = 3.53 \text{ seconds} $$ Substitute $t_{up}$ into vertical displacement: $$ y_{max} = h_0 + v_{0y} t_{up} - \frac{1}{2} g t_{up}^2 $$ Calculate each term: $$ v_{0y} t_{up} = 34.64 \times 3.53 = 122.2 $$ $$ \frac{1}{2} g t_{up}^2 = 0.5 \times 9.8 \times (3.53)^2 = 60.98 $$ Therefore: $$ y_{max} = 5 + 122.2 - 60.98 = 66.22 \text{ meters} $$ **Final answers:** - Time of flight: $\boxed{7.21}$ seconds - Maximum height above ground: $\boxed{66.22}$ meters