• 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`) 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`