Y^2 = Yo^2 + 2g*h = o
Yo^2 - 19.6*(150-50) = 0
Yo^2 = 1960
Yo = 44.27 m/s. = Ver. component of the
initial velocity.
Vo = Yo/sin A = 44.27/sin30 = 88.54 m/s.
= Initial velocity.
Xo = Vo*cos A = 88.54*cos30 = 76.68 m/s.
= Hor. component of initial velocity.
Y = Yo + g*Tr = 0 @ max. ht.
44.27 - 9.8*Tr = 0
9.8Tr = 44.27
Tr = 4.52 s. = Rise time.
h = 0.5g*t^2 = 150 m
4.9t^2 = 150
t^2 = 30.61
Tf = 5.53 s. = Fall time.
Range = Xo*(Tr+Tf) = 76.68*(4.52+5.53)=
770.6 m.
Suppose you are a general in the Napoleonic wars. You are no top of a plain 50 m high overlooking the enemy soldiers. You now have a brand new cannon that will decimate the enemy. Unfortunately in the christening process you used some rather cheap grade champagne and all the water in the champagne rusted the hinges of the cannon so that it is stuck angled upward at an angle of 30 from the horizontal. Your trusty new cannon expert insisted that the ball will reach a maximum height of 150m from the ground but unfortunately he was killed before he could tell you where the cannonball would land on the plain below. You have soldiers on the ground and you want to make sure that only the enemy soldiers are killed by your uber cannon. Determine the range of the cannon in order to keep your soldiers safe.
1 answer
1 answer