Question

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 slightly above it. Suppose the arrow speed is the same (70 m/s) and the target is 10 meters away. What launch angle, Θ, is needed so that the arrow hits the bulls-eye?

Answer 1

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

$\Delta y = 0$

$\Delta y = v_y t + \frac{1}{2} at^2$

$0 = v_y t - \frac{1}{2}9.8 t^2$

$t = \frac{2v_y}{9.8}$

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

so we will have

$R = v_x t$

$10 = v_x \frac{2 v_y}{g}$

now we know that

$v_x = v cos\theta$

$v_y = v sin\theta$

now we will have

$10 = \frac{2 vcos\theta vsin\theta}{g}$

$10 = \frac{70^2 sin2\theta}{9.8}$

$sin2\theta = 0.02$

$\theta = 0.57 degree$