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.
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Problem 1(a) Step 1: Maximum height formula for vertical throw upwards. v^2 = u^2 - 2gh At maximum height, v = 0, u = 20 m/s, g = 10 m/s^2. Step 2: Substitute values. 0 = (20)^2 - 2(10)h 0 = 400 - 20h Step 3: Simplify. 20h = 400 h = (400)/(20) = 20 m 20 m Problem 1(b) Step 1: Time to maximum height formula. v = u - gt At maximum height, v = 0. Step 2: Substitute values. 0 = 20 - 10t Step 3: Simplify. 10t = 20 t = 2 s 2 s Problem 1(c) Step 1: Total time to return to starting point is twice the time to max height (symmetric motion). T = 2t = 2 × 2 = 4 s 4 s Problem 1(d) Step 1: Velocity on return: magnitude same as initial, direction downward. v = -u = -20 m/s (downward) -20 m/s Problem 2(a) Step 1: Free fall from height h = 45 m, u = 0, g = 10 m/s^2. Velocity on ground. v^2 = u^2 + 2gh Step 2: Substitute values. v^2 = 0 + 2(10)(45) v^2 = 900 Step 3: Simplify. v = sqrt(900) = 30 m/s 30 m/s Problem 2(b) Step 1: Time of fall formula. h = ut + (1)/(2)gt^2 Step 2: Substitute values. 45 = 0 + (1)/(2)(10)t^2 45 = 5t^2 Step 3: Simplify. t^2 = 9 t = 3 s 3 s Problem 3 Step 1: Body falls freely for 2 s, then parachute opens, deceleration a = 2 m/s^2 until stop. Find total distance. First phase: free fall t_1 = 2 s, u=0, g=10 m/s^2. h_1 = (1)/(2)gt_1^2 = (1)/(2)(10)(2)^2 = 20 m v_1 = gt_1 = 10 × 2 = 20 m/s Step 2: Second phase: from v_1 = 20 m/s to v_2 = 0, a = -2 m/s^2. Time t_2. v_2 = v_1 + at_2 0 = 20 - 2t_2 t_2 = 10 s Step 3: Distance second phase. h_2 = v_1 t_2 + (1)/(2) a t_2^2 = 20(10) + (1)/(2)(-2)(10)^2 = 200 - 100 = 100 m Step 4: Total distance. h = h_1 + h_2 = 20 + 100 = 120 m 120 m Problem 4 Step 1: Elevator acceleration a = 2 m/s^2 upward, cable breaks, now a = -10 m/s^2. Was ascending at v_0 = 4 m/s. Time to stop relative to ground. Net acceleration now a = g downward regardless. But since ascending, time to stop. v = v_0 + at 0 = 4 - 10t t = 0.4 s 0.4 s Problem 5 Step 1: Train from rest, a = 3 m/s^2 for 30 s, then constant v. Distance in 1 min = 60 s. First phase: t_1 = 30 s. v = at_1 = 3 × 30 = 90 m/s s_1 = (1)/(2) a t_1^2 = (1)/(2)(3)(30)^2 = 1350 m Step 2: Second phase: t_2 = 30 s, constant v=90 m/s. s_2 = v t_2 = 90 × 30 = 2700 m Step 3: Total distance. s = s_1 + s_2 = 1350 + 2700 = 4050 m 4050 m