A 2.50 kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in a). c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

A) 46.43 m/s

B) Acceleration = 2395.27 m/s²

C) Average force in Newton's = 5988.18 N

Ratio of this force to weight = 244.14

Explanation:

We are given that;

Mass; m = 2.5 kg

Height reached by firework; y = 110m

Length of tube;L = 0.45m

A) According to kinematic equation;

v² = u² + 2gs

In this case, v = 0 and because gravity is acting against motion. Thus,

0 = u² - 2gy

u² = 2gy

u = √2gy

Plugging in the relevant values ;

u = √(2 x 9.8 x 110)

u = √2156

u = 46.43 m/s

B) we want to find the acceleration with u = 0 m/s and v = 46.43 m/s in the tube. The tube length is 0.45m

Thus,

v² = u² + 2aL

46.43² = 0 + (2•a•0.45)

2155.7449 = 0.9a

acceleration, a = 2155.7449/0.9 = 2395.27 m/s²

C) From Newton's law of motion,

Average Force; F = ma

Thus, F = 2.5 x 2395.27 = 5988.18 N

We are told to express it as a ratio to the weight of the shell.

Thus, ratio = F/W

Weight = mg and force = ma

Thus, F/W = ma/mg = a/g = 2395.27/9.8 = 244.42

a straight line sloped

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