1. **Problem statement:** A particle P is projected vertically upwards with speed 24 m/s from a point 5 m above ground level. We need to find the time from projection until P reaches the ground. 2. **Relevant formula:** Use the equation of motion under constant acceleration due to gravity $g = 9.8$ m/s$^2$ downward: $$ s = ut + \frac{1}{2}at^2 $$ where - $s$ is the displacement, - $u$ is the initial velocity, - $a$ is the acceleration, - $t$ is the time. 3. **Set up the problem:** Taking upward as positive, - Initial velocity $u = +24$ m/s, - Acceleration $a = -9.8$ m/s$^2$ (gravity acts downward), - Displacement $s = -5$ m (since the particle ends 5 m below the starting point, at ground level). 4. **Apply the equation:** $$ -5 = 24t - 4.9t^2 $$ Rearranged: $$ 4.9t^2 - 24t - 5 = 0 $$ 5. **Solve quadratic equation:** Using the quadratic formula: $$ t = \frac{24 \pm \sqrt{24^2 - 4 \times 4.9 \times (-5)}}{2 \times 4.9} $$ Calculate discriminant: $$ 24^2 = 576 $$ $$ 4 \times 4.9 \times (-5) = -98 $$ $$ \sqrt{576 + 98} = \sqrt{674} \approx 25.96 $$ 6. **Calculate roots:** $$ t = \frac{24 \pm 25.96}{9.8} $$ Two possible times: - $$ t = \frac{24 + 25.96}{9.8} = \frac{49.96}{9.8} \approx 5.10 \text{ s} $$ - $$ t = \frac{24 - 25.96}{9.8} = \frac{-1.96}{9.8} \approx -0.20 \text{ s (discard negative time)} $$ 7. **Final answer:** The time from projection until the particle reaches the ground is approximately **5.10 seconds**.