Question

In question 8 from “Problems 2”, you calculated that an archery arrow shot with an initial velocity of 45 m/s at an angle of 10 degrees would travel 70.67 m (2 d.p.) before hitting a target at the same height from which it was released. If the archery target was moved further away from the archer, discuss some kinematic-based strategies that the archer could adopt to ensure that the arrow reaches the target. Ensure that the variables discussed are relevant to your answer. Assume air resistance is negligible.

Answer 1

The distance traveled by the arrow and horizontal velocity are directly proportional, thus when the distance increases, an increase in initial velocity will allow the arrow to hit the target.

Time of motion of the projectile

If the horizontal distance the projectile would travel before hitting the target is 70.67 m, the time of motion of the projectile is calculated as;

X = Vx(t)

t = X/Vx

t = X/Vcosθ

t = (70.67) / (45 x cos10)

t = 1.6 s

When the archery target is moved further away from the archer, the archer needs increase the initial velocity of the arrow assuming the angle of projection is constant.

Since the distance and horizontal velocity are directly proportional to each other, thus when the distance increases, an increase in initial velocity will allow the arrow to hit the target.

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