Question

39. A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft above the ground. The crossbow fires her arrows at 300ft/s. (a) At what angle of elevation should she hold the cross- bow to hit her target? (b) If the deer is moving perpendicularly to her line of sight at a rate of 20mph, by approximately how much should she lead the deer in order to hit it in the de- sired location?

Answer 1

a)

$\theta$ = angle of elevation

v_o = speed of fire of arrow = 300 ft/s

consider the motion of arrow along the horizontal direction or X-direction

V_ox = initial velocity In X-direction = v_o Cos = (300) Cos

a_x = acceleration = 0

X = displacement = distance of the deer = 40 yards = 120 ft

t = time of travel = ?

using the equation

X = V_ox t + (0.5) a_x t²

120 = (300) Cos t + (0.5) (0) t²

t = 120/((300) Cos)                                                                          eq-1

consider the motion along the vertical direction or Y-direction

V_oy = initial velocity In Y-direction = v_o Sin = (300) Sin

a_y = acceleration = - 32.2 ft/s²

Y_o = initial position at the time of fire = height of the cross-bow = 5 ft

Y = final position at the time of hit = 4 ft

t = time of travel = 120/((300) Cos)

using the equation

Y = V_oy t + (0.5) a_yt²

4 = 5 + ((300) Sin) (120/((300) Cos)) + (0.5) (- 32.2) (120/((300) Cos))²

-1 = 120 tan - 2.576 sec²

-1 = 120 tan - 2.576 (1 + tan²)

tan = 0.013137

= 0.75 deg