an arrow is shot horizontally from the top of a tower at a speed of 15m/s and hits the ground with a speed of 25m/s. calculate the height of the tower

h = 20 m

Explanation:

Let's ignore air resistance.

The initial kinetic energy would be increased by the decrease in potential energy.

½m15² + m(9.8)h = ½m25²

h = (25² - 15²) / (2(9.8)) = 20.4081... ≈ 20 m

A. Final pressure P2

P2/P1 = (T2/T1)^n/n-1

P1 = 4bar

T1 = 438K

T2 = 300K

Polytropic index, n, = 1.3

P2 = 4 (300/438)^1.3/1.3-1

P2 = 4 (300/438)^4.333

P2 = 4 * 0.19400

P2 = 0.776bar.

B. The work done is;

W = mR/ n-1 (T1 -T2)

Where, R = 0.1889kJ/kg.K, m = 1

W = 1 * 0.1889/ 1.3-1 * (438-300)

W = 86.89kJ/kg.

C. The heat transfer, Q

Q = W + ΔU

Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk

Q = 86.89 + 1 * 0.743 (300-438)

Q = 86.89 + (-102.534)

Q = -15.644kJ/K

Q = 15.64kJ/K