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An apple is dropped from a height of 15 meters above the ground. An arrow is shot at the same moment at the apple from the ground a horizontal

distance of 100 meters away. Show that if the initial velocity vector is in the direction of the apple and the arrow will travel at least 100 meters that the arrow will hit the apple. use g=10 m/s^2



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Angle of projection of Arrow = Ф

tan Ф = 15 / 100    => 100 tan Ф = 15

The initial velocity u of arrow is such that it travels at least 100m.  
Time taken for the arrow to travel 100 meters =  t₁ = 100 / (u Cos Ф)

Hence at t = t₁, both arrow and apple are at the same horizontal position.

Vertical height of arrow at time t₁ = u Sin Ф t₁ - 1/2 g t₁²
                   = u Sin Ф ( 100/ u Cos Φ) - 1/2 g t₁²
                   = 100 tan Φ - 1/2 g t₁² 
                   = 15 - 1/2 g t₁²

In time t1,  the Apple drops by a distance of 1/2 g t₁² and its vertical position above ground is =  15 - 1/2 g t₁² .

So both heights are exactly same at t = t₁ = 100 / u Cos Ф.

Thus the Arrow hits the apple.
To Find the minimum speed of Arrow to hit the apple:

Range of arrow = R = u² sin 2Ф /g   >= 100 m  for the arrow to hit apple.

u² >= 100 g / Sin 2Ф 
     >= 1, 000 / [ 2 tanФ / (1+tan²Ф) ] = 5 * 409 / 6 
u   >=  18.46 m/sec


The apple and the arrow have the same acceleration due to gravity.  There is no acceleration in the horizontal direction.

Hence, from the frame of reference of Arrow, the apple is stationary.  Initially it is aimed at the Apple with the speed = u, in the direction of apple.  Hence it will hit the apple after time =  distance/ speed 
             = [ 100/Cos Ф ] / u  = 100 / (u Cos Ф).

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