Question

A 0.20 kg ball is held stationary by a jet of water from falling. At the nozzle, the speed of the jet is 5m/s, and its diameter is 10mm. Find the velocity at which the water jet hits the ball and the height of the ball. Density of water is 1000kg/m2. Assume there is no change in the diameter of the water jet. 2.1 Find the velocity of the jet as it hits the ball (hint: the weight of the ball is held up by the force of the jet of water) 2.2 Find the height at which the ball is held

Answer 1

0 Assumption & a) No change in diameter of Jet of water 6) Water upon striking ball with negative velocity. reverts back completely h = ? Jet or Stream w ball = m = 0.2 kg Pw = density of water dia=constant = 1000 kg/m3 d= diameter of nozzle exit = 10mm of water h=0 1 g = 9.81 m/ 2 - Nozzle Vaug = 5 m/s a Force on ball must be net = 0 So that it is in Mechanical Equilibrium Thus Force of Gravitation = Force due to water striking ball. 0.2 kg 4 (918) +) = (-m w: 1) - (m_vert) where st m = mass striking in at time Newton's 2nd law esta velocity of striking ball. & -mut) 1 = final momentim after striking ball.

(must) 1 = initiall promentim before striking ball. 0.2x9.81 = 1-2mUst I = 2 in Ust st where in rate at which jet - Pu * Ad x ang ef water strokes ball = M = 10' x 4x(10-2)** 5 = 045*. 0.2 x 9.81 - 2 x (m = 0.52 - Vst 0.2 x 9.81 x 2 x 2 = 2-5 Am velocity with which = &st jet of water strikes ball = 2.5 m/s - where; 6) Using [2² - u² = -2gh & Vt = 2.5 m/s u= 5 m/ Eogh of motion S = height of ball 2.52 - 5² = - 2x9.81xh E{ g = 9.81 m/s2 h = height of ball = 0.956 m from nozzle