An archer lets go of an arrow aimed at a target with a velocity of $65.0\,\text{m s}^{-1}$ at an angle of $4.30^{\circ}$ above the horizontal, as illustrated in Fig. 2.1. At the moment of release, the tip of the arrow is $70.0\,\text{m}$ horizontally from the target and $1.66\,\text{m}$ above the horizontal ground. The arrow strikes the centre of the target. Take air resistance to be negligible and assume that the entire mass of the arrow is concentrated at its tip.
(a)[2]
Show that the arrow takes $1.08\,\text{s}$ to arrive at the target.
(b)[3]
Calculate the height of the target’s centre above the ground.
(c)[2]
Using the energy changes, state and explain how the arrow’s final kinetic energy as it reaches the target compares with its initial kinetic energy just after release. No numerical calculation is needed.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Applying $t = \frac{s}{v}$” …