1. **Problem statement:** A bullet is fired upward at an angle of 60° with an initial velocity of 100 m/s from a tower 100 m high. We need to find: - Maximum height above the ground - Total time of flight - Velocity on landing - Horizontal distance from the base of the tower 2. **Given data:** - Initial velocity $u = 100$ m/s - Angle of projection $\theta = 60^\circ$ - Height of tower $h = 100$ m - Acceleration due to gravity $g = 9.8$ m/s$^2$ 3. **Resolve initial velocity into components:** $$u_x = u \cos \theta = 100 \times \cos 60^\circ = 100 \times 0.5 = 50 \text{ m/s}$$ $$u_y = u \sin \theta = 100 \times \sin 60^\circ = 100 \times \frac{\sqrt{3}}{2} = 86.6 \text{ m/s}$$ 4. **Maximum height above the launch point:** Use formula for vertical motion: $$v_y^2 = u_y^2 - 2 g s$$ At max height, vertical velocity $v_y = 0$, so $$0 = u_y^2 - 2 g s \implies s = \frac{u_y^2}{2g} = \frac{(86.6)^2}{2 \times 9.8} = \frac{7500}{19.6} = 382.65 \text{ m}$$ 5. **Maximum height above the ground:** Add tower height: $$H = h + s = 100 + 382.65 = 482.65 \text{ m}$$ 6. **Total time of flight:** Use vertical motion equation for displacement from initial height to ground (displacement $= -100$ m): $$y = u_y t - \frac{1}{2} g t^2$$ Set $y = -100$: $$-100 = 86.6 t - 4.9 t^2$$ Rearranged: $$4.9 t^2 - 86.6 t - 100 = 0$$ Use quadratic formula: $$t = \frac{86.6 \pm \sqrt{(86.6)^2 + 4 \times 4.9 \times 100}}{2 \times 4.9}$$ Calculate discriminant: $$86.6^2 + 1960 = 7500 + 1960 = 9460$$ $$\sqrt{9460} = 97.26$$ So, $$t = \frac{86.6 + 97.26}{9.8} = \frac{183.86}{9.8} = 18.77 \text{ s}$$ (Discard negative root) 7. **Velocity on landing:** Horizontal velocity remains constant: $$v_x = 50 \text{ m/s}$$ Vertical velocity at landing: $$v_y = u_y - g t = 86.6 - 9.8 \times 18.77 = 86.6 - 183.95 = -97.35 \text{ m/s}$$ Magnitude of velocity: $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{50^2 + (-97.35)^2} = \sqrt{2500 + 9476} = \sqrt{11976} = 109.44 \text{ m/s}$$ Direction angle below horizontal: $$\phi = \tan^{-1} \left( \frac{|v_y|}{v_x} \right) = \tan^{-1} \left( \frac{97.35}{50} \right) = 62.3^\circ$$ 8. **Horizontal distance from base:** $$x = u_x t = 50 \times 18.77 = 938.5 \text{ m}$$ **Final answers:** - Maximum height above ground: $482.65$ m - Total time of flight: $18.77$ s - Velocity on landing: $109.44$ m/s at $62.3^\circ$ below horizontal - Horizontal distance from base: $938.5$ m