1. The figure below illustrates the trajectory of a cannonball. Assume that the cannonball was fired into the air with an initial speed of vo = 100m/s at position 0. The cannonball landed at position 2 that is at a horizontal distance 1 = 1000m measured from position 0. Calculate the angle of takeoff 0, time of flight t2, and maximum height h, that the cannonball reached. Note that 2 sin 0 cos 0 = sin (20). For simplicity, take g = 10 m/s². x = xo + (Vo Cos 0)t 1 y = yo + (Vo sin 0)t Vx = Vo Cos e Vy = Vo sin e - gt
1. The figure below illustrates the trajectory of a cannonball. Assume that the cannonball was fired into the air with an initial speed of vo = 100m/s at position 0. The cannonball landed at position 2 that is at a horizontal distance 1 = 1000m measured from position 0. Calculate the angle of takeoff 0, time of flight t2, and maximum height h, that the cannonball reached. Note that 2 sin 0 cos 0 = sin (20). For simplicity, take g = 10 m/s². x = xo + (Vo Cos 0)t 1 y = yo + (Vo sin 0)t Vx = Vo Cos e Vy = Vo sin e - gt
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Transcribed Image Text:1. The figure below illustrates the trajectory of a cannonball. Assume that the cannonball was fired into
the air with an initial speed of vo = 100m/s at position 0. The cannonball landed at position 2 that is at
a horizontal distance 1= 1000m measured from position 0.
Calculate the angle of takeoff 0, time of flight t2, and maximum height h, that the cannonball reached.
Note that 2 sin 0 cos 0 = sin (20).
For simplicity, take g = 10 m/s².
x = x0 + (Vo cos 0)t
2)
1
y = yo + (o sin 0)t –
Vx = Vo Cos O
Vy = Vo sin e -gt
%3D
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