Question

for his second try, robin decides to change his position but release the arrow using the same speed and angle as his first try. assume that the orange remains in the same position. how far from the orange should robin stand in order to hit the orange? there may be more than one correct distance. 13.7 m 101 m 6.4 m 108 m 61.3 m

Explanation:

To solve this, we analyze the projectile motion of the arrow. When the speed and angle of projection are constant, the range (horizontal distance to hit the target) depends on the initial velocity components and gravitational acceleration. For a projectile, the range formula is \( R=\frac{v_0^2\sin(2\theta)}{g} \), but also, for a given speed and angle, there can be two possible ranges (if the target is at a height such that the projectile can hit it on the way up or down), but in typical cases with a target at the same height (orange's position), the valid ranges should be reasonable for an arrow's shot.

• 101 m and 108 m are too large for an arrow's range with typical speeds.
• 6.4 m and 13.7 m could be too short, but 61.3 m is a more reasonable range for an arrow shot with appropriate speed and angle. Also, sometimes projectiles can have two ranges, but among the options, 13.7 m and 61.3 m (or others) – but typically, for such problems, the correct distances are the ones that fit the projectile range. Assuming the first try had a certain range, and now with same speed and angle, the valid distances (from the options) that make sense for arrow's range are 13.7 m and 61.3 m (as projectiles can have two ranges: one shorter (nearer) and one longer (farther) when the target is at the same height, due to the parabolic path – hitting on the way up or down). But let's check the options:

Looking at typical projectile range for an arrow (assuming initial speed around, say, 50 m/s and angle 30 degrees: \( R=\frac{50^2\sin(60^\circ)}{9.8}\approx\frac{2500\times0.866}{9.8}\approx222 \), but that's too big. Wait, maybe lower speed. If speed is around 30 m/s, angle 30 degrees: \( R=\frac{30^2\sin(60^\circ)}{9.8}\approx\frac{900\times0.866}{9.8}\approx79.7 \). So 61.3 m is close. Also, if angle is steeper, shorter range. So 13.7 m (shorter, steeper angle) and 61.3 m (longer, shallower angle) could be valid. But the options given: 13.7 m, 61.3 m are the ones that make sense (101, 108 too long; 6.4 too short).

The problem involves projectile motion (Physics subfield of Natural Science). For a projectile with constant speed and angle, the horizontal distance (range) to hit a target at the same height can have two valid values (hitting on the way up or down). Among the options, 13.7 m (shorter range, steeper angle) and 61.3 m (longer range, shallower angle) are reasonable for an arrow’s shot, while 6.4 m (too short), 101 m, and 108 m (too long) are not.

Answer:

13.7 m, 61.3 m