Ahorizontal jet of water is made to hit a vertical wall with a negligible rebound. if the speed of water from the jet is 'v', the diameter of the jet is 'd' and the density of water is 'p' , then what is the force exerted on the wall by the jet of water?

the answer is a 3.8cm

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft