This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: For maximum height, final velocity v=0 m/s. v^2 = u^2 - 2gh Substitute u=20 m/s, g=10 m/s^2: 0 = (20)^2 - 2 × 10 × h 400 = 20h h = 20\ m 20\ m प्रश्न (2) Step 1: Take upward as positive direction from top of tower. Displacement to ground s=-50 m, u=20 m/s, a=-g=-10 m/s^2. s = ut + (1)/(2)at^2 Substitute values: -50 = 20t + (1)/(2)(-10)t^2 -50 = 20t - 5t^2 Step 2: Rearrange into standard quadratic equation. 5t^2 - 20t - 50 = 0 Divide by 5: t^2 - 4t - 10 = 0 Step 3: Solve using quadratic formula t = -b ± sqrt(b^2 - 4ac)2a, where a=1, b=-4, c=-10. t = 4 ± sqrt((-4)^2 - 4(1)(-10))2(1) t = 4 ± sqrt(16 + 40)2 = 4 ± sqrt(56)2 = 4 ± 2sqrt(14)2 = 2 ± sqrt(14)\ s Step 4: Take positive root (time cannot be negative): t = 2 + sqrt(14) s. 2 + sqrt(14)\ s प्रश्न (3) Step 1: Time to reach maximum height t = (u)/(g). Substitute u=19.6 m/s, g=9.8 m/s^2: t = (19.6)/(9.8) = 2\ s 2\ s प्रश्न (4) Step 1: Time to fall from height 50 m with initial vertical velocity 0 m/s. h = (1)/(2)gt^2 50 = (1)/(2) × 10 × t^2 50 = 5t^2 t^2 = 10 t = sqrt(10)\ s Step 2: Horizontal distance (range) R = u_x t, where horizontal velocity u_x=10 m/s (constant). R = 10 × sqrt(10) = 10sqrt(10)\ m 10sqrt(10)\ m