Question

A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole 12 m deep. In other words, assume that the rock lands 12 m lower than the height from which it was thrown. Take "up" to be the positive direction for this problem. Although the numbers in this problem have less than 3 significant figures, please enter your answers with at least 3 significant figures, and remember that velocity is a vector. (a) What is the rock's velocity as it hits the bottom of the hole? Incorrect: Your answer is incorrect. m/s (b) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Answer 1

initial speed (u) = 20m/s. final speed cv) = 0 m/s The Reight it will go from initial point VR-u² = zas =) ha u2- 20x20 = 20.4082m. 2x9.8 9 9.8 . time Ctid- . 2 2.040829. (a) from the highest point it has to cover h 20.4088+12= 32.4087m. initial velocity ced=0 v2 = uztas - 20h = v= I lagh = ± 25.203189 m/s fore downward V=-25.2 m/s t = 2.5718 8 , (6) ta titt = 4.618 Ang Please reate dear. 5.2 m/s Ang