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 + (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`