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*.

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*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

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*Alternately,*

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).