Question

An open cylindrical tank of acid rests at the edge of a table 2.4 m above the floor of the chemistry lab.

Part A

If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is 55 cm deep?

PLease show all your work.

Answer 1

By Torricelli's law, speed of the acid in horizontal direction from the small hole of tank is defined as V_H = squareroot 2 g d Here d = depth of the hole Initial vertical velocity = (V_y)_i = 0 consider it take the time t. when acid hit to the floor then by equation of kinetics. H = (v_y)_i t + 1/2 gt^2 h = 0 times t + 1/2 gt^2 t = squareroot 2h/g Now, velocity along horizontal direction (V_H) will not change. It means it will be uniform because there is no any acceleration along horizontal direction. Then, Horizontal distance from the bottom of the table where acis hit to the floor = x = V_H t X = squareroot 2 g d times squareroot 2h/g = squareroot 2gd times 2h/g = squareroot 4 dh X = 2 squareroot dh let us put all these values