Question

of 12.0 m/s from the edge of a cliff 70.0 m high. What is the magnitude of its speed just before hitting the ground at the base of the clif? (answer should look A stone is thrown vertically upward with a speed like 18.9 m's) 120

Answer 1

The ball is thrown from a height, h = 70 m Let, the mass of the ball = m Initial velocity of the ball, v = 12.0 m/s The total energy of the ball at the initial condition, E_tot = E_ini = E_ki + E_P_i = (mgh + 1/2 mv^2) Let, the magnitude of the speed of the ball just before hitting the ground at the base of the cliff = v_f. Energy of the ball just before hitting the ground at the base of the cliff, E_fin E_k_f + E_k_f + E_p_f = 1/2 mv^2_f + 0 = 1/2 mv^2_f. As there is no loss of energy in the initial & find state, E_fin E_ini = > 1/2 mv^2_f = 1/2 mv^2 + mgh = > v^2_f = (v^2 + 2gh) v_f = squareroot v^2 + 2gh = squareroot (12)^2 + 2(9.81) times 70 = 38.954 m/s