A battleship fires a cannon to hit a pirate ship with velocity 200 m/s and angle with horizontal direction 37o.
• Find:
• A) the position of the bullet after 2 secs, velocity and acceleration.
• B) the highest altitude it will reach and time it takes to get there
• C) Find the range of the cannon and see if it will hit the pirate ship (the distance between the battleship and pirate ship is 5 km)
• D) what should be the angle of the cannon with horizontal to hit the pirate ship.
Vo = 200m/s[37o].
Xo = 200*Cos37 = 160 m/s.
Yo = 200*sin37 = 120 m/s.
A. Y = Yo + g*t = 120 + (-9.8)*2 = 100 m/s. = Ver. component of velocity @ 2 s.
Tan A = Y/Xo = 100/160 = 0.625. A = 32o.
V = Xo/Cos A = 160/Cos32 = 189 m/s[32o].
B. Yf = Yo + g*Tr
Yf = 0, g = -9.8m/s^2, Tr = Rise time,
Tr = ?.
h = Yo*Tr + 0.5g*Tr^2 = Max ht.
C. Range = Vo^2*sin(2A)/g Meters.
It will be less than 5 km.
D. The 5km range cannot be met with the
given velocity of 200 m/s. With a given velocity, the max range occurs at 45o.
The required velocity is 221 m/s:
Range = Vo^2*sin(2A)/g = 5,000 m.
Vo^2*sin(90)/9.8 = 5,000.
Vo^2*1/9.8 = 5000.
Vo^2 = 49000.
Vo = 221 m/s.