A 1500 kg cannon is fired from an upper deck of a castle. When the cannon fires

Question

A 1500 kg cannon is fired from an upper deck of a castle. When the cannon fires a 16 kg cannon ball horizontally, the cannon rolls backward with an initial recoil speed of 4 m/s. (a) Find the cannon's muzzle velocity (speed with which the cannon ball exits the cannon). Assuming this muzzle velocity is the same for each launch, if the cannon is tilted to 20 degree above the horizontal, find: b) the cannon's new recoil speed (c) the maximum height and range the cannon ball reaches, assuming the muzzle of the cannon is 25 m above the ground level.

Explanation / Answer

(A) Applying momentum conservation for the firing in horizontal direction,

cannon is fired horizontally,

so 0 = (16 v0) - (1500 x 4)

v0 = 375 m/s

(B) when cannon tilted above 20 deg,

0 = (16 v0 cos20) - (1500 x 4)

v0 = 399 m/s

(C) at maximum height, vy = 0

vy^2 - v0y^2 =2 g H

0^2 - (399 sin20)^2 = 2 x -9.81 x H

H = 949.5 m   

so maximum height from ground, Hmax = H + 25 = 974.5 m .......Ans

as cannon reaches ground,

y = v0y t + ay t^2 /2

- 25 = (399 sin20)t - 9.81 t^2 /2

4.905 t^2 - 136.5t - 25 = 0

t = 28 sec

Range = (v0x ) (t)

= (399 cos20) (280)

= 104982 m Or 104.98 km

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