Question

An arrow is shot in an arc to hit a target that is 5.2 m above the ground. If the arrow leaves the bow 0.89 m above the ground travelling 46 m/s, how fast is it travelling when it hits the target?

Answer 1

The law of conservation of energy says

$$\frac{mv_1^2}{2}+mgh_1=\frac{mv_2^2}{2}+mgh_2$$

Hence, the final speed of an arrow

$$v_2=\sqrt{v_1^2+2g(h_1-h_2)}\\ =\sqrt{46^2+2*9.8*(0.89-5.2)}=45\:\rm m/s$$