Question

2. A water balloon is thrown from the roof vertically upwards with initial speed o. It reaches some maximum height Hea then falls, and hits the ground 25 meters below the roof with the speed 2 times faster than the initial speed. (a) What is the initial speed Vy? (b) What is the speed it hits the ground? (c) What is the maximum height Ho? (d) How long does it take to reach the Ho?(e) How much total time does it take to reach the ground?

Answer 1

v^2 - u^2 = 2as [Ballon moving upward) v(final velocity) = 0 -v^_0 = 2(-g) H H_max = v^2_0/2g v = u + at = > t_1 = v_c/g s = ut + 1/2 at^2 [from B to C] [H + 25] = 1/2 g t^2 = > t^2_2 = 2(H + 25]/g v^2 - u^2 = 2as = > (2v_0)^2 = 2 times g times [H + 25] = > v^2_0 = g/2 (H + 25) Putting (4) in (1) H = g/2 (H + 25)/(2g) = > 4H = H + 25 = > 3H = 25 H = 25/3 = 8.33m eqn (1) v^2_0 = 2gH = 2 times 8.33 = > v^2_0 = 166.66 = > v_0 = 12.91 m/s 2v_0 = 2 times 12.91 = 25.82m/s H = 8.33m = > h = 33.33m t_1 = v_0/g = 12.91/9.8 = 1.32 sec = > t_1 = 1.32 sec = u + at = > v_f/a = t = > t = 2 v_0/g [from B to C] u = 0 t = 2.634 sec