## Freely Falling Objects

• free-fall acceleration: g • magnitude of g: 9.80\ m//s^2 • direction of acceleration: downward or (a_y = -g=-9.80\ m//s^2)
v_f = v_i + g t v_(avg) = (v_i + v_f)/2 y_f = y_i + 1/2(v_i + v_f)t y_f = y_i + v_it + 1/2 g t^2 v_f^2 = v_i^2 + 2g(y_f - y_i)
Example: A stone thrown from the top of a building is given an initial velocity of 20.0\ m//s straight upward. The stone is launched 50.0\ m above the ground, and the stone just misses the edge of the roof on its way down as shown in the Figure. (A) Using t_A= 0 as the time the stone leaves the thrower’s hand at position A, determine the time at which the stone reaches its maximum height. Given: v_i = 20.0\ m//s g = -9.80\ m//s v_f = 0 Use Equation: v_f = v_i + g t => t = (v_f - v_i)/g =>t=(0-20.0)/9.80 = 20.4 s (B) Find the maximum height of the stone. set y_i = y_A=0 y_f = y_(max) = y_B y_B = y_A +v_it + 1/2 g t^2 y_B = 0 + (20.0\ m//s)(2.04\ s) + 1/2(-9.80\ m//s^2)(2.04\ s)^2 y_B = 20.4\ m (C) Determine the velocity of the stone when it returns to the height from which it was thrown. Know:Y_A = Y_C =0 v_C^2 = V_A^2 + 2a(Y_C-Y_A) Know:Y_A = Y_C =0 =>v_C^2 = V_A^2 => v_C = ±V_A since V_C and V_A are opposite, => V_C = - V_A = -20.0\ m//s (D) Find the velocity and position of the stone at t = 5.00\ s. v_D = v_i + g t = 20.0\ m//s + (-9.80\ m//s^2)(5.00\ s) =-29.0\ m//s y_D = y_i + v_it + 1/2g t^2 = 0 + (20.0\ m//s)(5.00\ s) +1/2(-9.80\ m//s^2)(5.00\ s) =-22.5 m