Question

A 0.2 kg cannon ball is fired at an upward angle of 45° from the top of a 165 m cliff with a speed of 175 m/s. (A) Using conservation of energy, find the speed it has when it strikes the ground below. Suppose the cannon ball lands in a soft, muddy field. The force of the field is 75 N on the projectile. (B) How deep does it penetrate into the ground?

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Answer 1

m = mass of cannonball = 0.2 kg

$\theta$ = angle from horizontal = 45

h = height = 165 m

V = speed at top = 175 m/s

V' = speed at bottom

A)

using conservation of energy

Potential energy at Top + Kinetic energy at top = Kinetic energy at bottom + Potential energy at bottom

mgh + (0.5) m V² = (0.5) m V'² + mg (0)

V'² = V² + 2gh

V'² = 175² + 2 x 9.8 x 165

V' = 184.01 m/s

B)

a = retardation caused inside the ground = - F/m = - 75 / 0.2 = - 375 m/s²

V_f = final speed = 0

d = displacement inside the ground

using the equation

V_f² = V'² + 2 a d

0² = 184.01² + 2 (-375) d

d = 45.2 m

m = 5 lb = 2.27 kg

h = 1 m

energy = potential energy gained = mgh = 2.27 x 1 x 9.8 = 22.25 J