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Sample 2: RJC Prelim 1997 Question 1(a)

A tank is on a flat plain with hills over 1000m away. The tank commander spots a target at a horizontal distance of 1825m and at a height of 200m up the side of a hill. The tank accelerates from rest towards the target at 0.5ms^-2 and after travelling 30 seconds fires a shell at the target.

(i) By considering the initial horizontal (u) and vertical (w) components of the shell's velocity relative to the ground, show that the muzzle velocity (V - the velocity relative to the gun) is related to the elevation of the gun (a) by:

V = (u - 15)/cosa = w/sina

(ii) Show that in order for the target to be hit the initial horizontal and vertical components of velocity of the shell must be related by the equation: 8uw = u² + 6400g, where g is the acceleration due to gravity.

Thinking Process:

Part (i)

Step 1:

First, recognise that the information "hills over 1000m away" is a red herring which is irrelevant to the question.

Step 2:

Using the equation v = u + at, determine the horizontal velocity of the tank after 30 seconds:

v = 0 + (0.5)(30) = 15 ms^-1

Step 3:

Since the tank has a horizontal velocity of 15ms^-1 , the horizontal velocity of the shell relative to the gun will be (u - 15)ms^-1

Since the tank has no vertical velocity, the vertical velocity of the shell relative to the gun is still w ms^-1

Step 4:

Resolving the velocity V into horizontal and vertical components,

V_x = (u -15) = Vcosa

V_y = w = Vsina

Hence = V = (u - 15)/cosa = w/sina.

Part (ii)

Step 1:

Calculate how far the target is from the tank, after the tank has accelerated for 30 seconds, using s = ut + ˝at²

s = 0(30) + ˝(0.5)(30)² = 225m

Step 2:

Since the target was initially 1825m from the tank, after 30 seconds, the distance has shrunk to 1825 - 225 = 1600m.

Step 3:

In order for the shell to hit the target, its horizontal displacement must therefore be 1600m at the instant when its vertical displacement reaches 200m (because target is 200m above the ground).

Step 4:

Apply the equation s = ut + ˝at² again to both the horizontal and vertical components:

Horizontally,

1600 = ut + ˝(0)t²

t = 1600/u ........ (1)

(since horizontal acceleration is zero)

Vertically,

200 = wt - ˝gt² ........ (2)

(g is negative because acceleration due to gravity is downwards, opposite to direction of velocity)

Step 5:

Substituting equation (1) into equation (2),

200 = w(1600/u) - ˝g(1600/u)²

200 = 1600w/u - 1280000g/u²

1 = 8w/u - 6400g/u²

u² = 8wu - 6400g

8wu = u² + 6400g

Q.E.D