A 3500 g fireworks shell is fired straight up from a mortar and reaches a height of 210.0m. (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 0.350 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 nutshell in the mortar? Express you answer in newtons and as a ratio to the weight of the shell?
kinetic energy at the ground has to be enough for m g h at 210 meters up
(1/2) m v^2 = m g h
v^2 = 2 g h = 2 * 9.81 * 210 = 4120
v = 64.2 meters/ second
a = change in velocity/ change in time = (64.2-0)/time
to get time average sped in tube = 64.2/2 = 32.1 m/s
time = 0.350 / 32.1 = 0.0109 seconds
so
a = 64.2 / .0109 = 5890 m/s^2
F = m a = 3.500 * 5890 = 20,600 Newtons (yikes)
ratio = 20,600 / (3.5*9.81) = about 600 times the weight