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Cannons and projectiles

A 30 kg object, A, was fired from a cannon in projectile motion. When the projectile was at its maximum height of 25 m, its speed was 20 m/s. An identical object, B, was attached to a mechanical arm and moved at a constant speed of 20 m/s in a vertical half-circle. The length of the arm was 25 m.

Ignore air resistance.
(a) Calculate the force acting on object A at its maximum height.
(b) Calculate the time it would take object A to reach the ground from its position of maximum height.
(c) Describe and compare the vertical forces acting on objects A and B at their maximum heights.


(a) Force on projectile = mg = 30x9.81 = 294.3 N
(b) Maximum height = 25 m using s = ut + at2 and with the initial vertical speed at the top u = 0 we have:
25 = gt2 therefore t = [50/9.8]1/2 = 2.25 s
(c) The object on the mechanical arm is subject to a centripetal force to keep it moving in a circle. This force (F) is given by the equation:
F = mv2/r = [30x400]/25 = 480 N which is greater than the weight of the object. Some of the force will be provided by the weight of the object.
The additional force of the arm on the object needed to keep it moving in a circle is therefore 480 - 294.3 = 185.7 N

I have worked out the angle of projection as well (you don't actually need that!)
At maximum height its vertical speed is zero (if it wasn't it wouldn't be the maximum height).
The horizontal speed is constant and equal to 20 m/s.
Take the initial velocity to be v at an angle to the horizontal of A
and so 20 = v cos A 400 = v2cos2A
Vertically: initial velocity = vsin A and 0 = v2sin2A 2x9.81x25
Combining the horizontal and vertical equations:

400tan2A = 2x9.8x25 this gives A = 47.9o

© Keith Gibbs 2016