Question

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 68.0 m away, making a 3.00o angle with the ground. How fast was the arrow shot?

Answer 1

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

$Angle=3^{\circ}$

as the arrow travels it acquire a vertical velocity $v_y$

$v_y=u+at$

$v_y=0+9.81\times t$-------1

Range is given by

R=ut

where u=initial velocity

$68=u\times t$

$t=\frac{68}{u}$

substitute the value of t in eqn 1

$v_y=9.81\times \frac{68}{u}$

$v_y\times u=9.81\times 68=667.08$--------2

and $tan(3)=\frac{v_y}{u}$

$v_y=utan(3)=0.0524u$

substitute it in 2

$0.0524 u^2=667.08$

$u^2=12,728.644$

u=112.82 m/s