If you were to aim right at the bulls-eye, the arrow would fall under gravity and drop below the intended target. If you intend to hit a bulls-eye it is necessary that you aim s…

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asked Aug 10, 2019 172k views

1 Answer

Mkk · Answer 1 · 2019-08-15T23:17:02+0000

Here range of the arrow where it will hit the target is given as

R = 10 m

now the vertical displacement of the arrow for whole motion is ZERO

so we will have

$\Delta y = 0$

$\Delta y = v_y t + (1)/(2) at^2$

0 = v_y t - (1)/(2)9.8 t^2

t = (2v_y)/(9.8)

now in the above time it will hit the target at horizontal distance of R = 10 m

so we will have

R = v_x t

10 = v_x (2 v_y)/(g)

now we know that

$v_x = v cos\theta$

$v_y = v sin\theta$

now we will have

$10 = (2 vcos\theta vsin\theta)/(g)$

$10 = (70^2 sin2\theta)/(9.8)$

$sin2\theta = 0.02$

$\theta = 0.57 degree$