Question

b. A girl, standing on the roof of a house which is 5.00 m high, throws a stone vertic upward. It strikes the ground 7.00 s after being thrown. Calculate: The initial velocity of the stone The maximum height the stone reaches above a. b. i. The roof ii. The ground and c. The velocity of the stone as it strikes the ground

Answer 1

a) Let v_0 be the initial speed of the stone. From point A to point B from first law of motion. v = u + at where, v = 0 (final speed, at maximum height) a = -g (gravitational acceleration) t = t_1 and u = v_0 (initial speed) 0 = v_0 - gt_1 t_1 = v_0/g \r\n Now from third law of motion in A to B v^2 = u^2 + 2as 0 = v^2_0 - 2gs s = v^2_0/2g (i) Now from B to c from second law of motion s + 5 = 0 times t_2 + 1/2 gt^2_2 s = 5 = 1/2 gt^2_2 v^2_0/2g + 5 = 1/2 gt^2_2 \r\n t^2_2 = v^2_0 + 10g/g^2 t_2 = squareroot v^2_0 + 10g/g given that t_1 + t_2 = 7 Now we put the value of t_1 & t_2 v_0/g + squareroot v^2_0 + 10g/g = 7 v_0 + squareroot v^2_0 + 10g = 7g squareroot v^2_0 + 10g = 7g - v_0