Answer to Question #269392 in Physics for Shihab

Question

A blast pump is pumping 5000 tons of water in one hour. A leakage occurs in the supply pipe and stream of water is coming out in a vertical direction with an initial velocity of 45 m/s and takes 3.5 seconds to reach its maximum height.

(a) Calculate the maximum height that water jet will reach.

(b) Draw the velocity - time graph for vertical motion of the water jet and with the help of graph justify the answer of part (a).

(c) With what velocity the water will touch the ground when it returns back om its maximum height?

Answer 1

In the calculations we neglect air resistance.

(a) $h=v^2/(2g)=45^2/(2\cdot9.8)=103.31\ (m)$.

In real conditions, the height will be much lower.

(b)

Under the condition of the problem, the jet reaches its maximum height at the end of 3.5 s. Ignoring the air resistance, we obtained a maximum height of 103.3 m, which corresponds to the motion of the jet for 4.6 s.

(c) $\rho gh=\rho v^2/2\to v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot103.31}=45\ (m/s)$