Question

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull's-eye if its initial speed is 37 m/s?

Answer 1

`16.25^(\circ)`

Step-by-step explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

Horizontal range is given by

`R=(v^2\sin2\theta)/(g)\\\Rightarrow \theta=(\sin^(-1)(Rg)/(v^2))/(2)\\\Rightarrow \theta=(\sin^(-1)(75* 9.81)/(37^2))/(2)\\\Rightarrow \theta=16.25^(\circ)`

The angle at which the arrow is to be released is
`16.25^(\circ)`.