Question

please help

A marble rolls off a tabletop 1.3 m high and hits the floor at a point 2.2 m away from the table's edge in the horizontal direction (a) How long (in s) is the marble in the air? (b) What is the speed of the marble (in m/s) when it leaves the table's edge? m/s (c) What is its speed (in m/s) when it hits the floor?

Answer 1

Given H = 1.3m R = 2.2m u_y = 0 using 3rd equation of motion y = u_yt + 1/2 a_y t^2 H = 1/2 times g times t^2 t = squareroot 2H/g = squareroot 2 times 1.3/9.8 t = 0.514 s Now R = u_xt + 1/2 a_xt^2 a_x = 0 2.2 = u_x times 0.514 v_x = u_x = 4.28 m/s v_y = 0 + 9.8 times 0.514 v_y = 5.037 m/s u = squareroot v_x^2 + v_y^2 = squareroot 18.31 + 25.37 = 43.68 v = 6.609 m/s