To determine the displacement of the arrow at 10.0 seconds after launching, we need to analyze the motion of the arrow separately during its ascent and descent phases.
During the ascent phase, the arrow is moving against the force of gravity, so its velocity decreases until it reaches its maximum height and momentarily comes to rest. Then, during the descent phase, the arrow accelerates downward due to the force of gravity.
Let's calculate the displacement for each phase:
- Ascent phase: We can use the kinematic equation for displacement in vertical motion: Δy = v₀t - (1/2)gt²
Here: v₀ = 39.2 m/s (initial velocity) t = 10.0 s (time) g = 9.8 m/s² (acceleration due to gravity)
Δy = (39.2 m/s)(10.0 s) - (1/2)(9.8 m/s²)(10.0 s)² = 392 m - (1/2)(9.8 m/s²)(100 s²) = 392 m - (1/2)(9.8 m/s²)(100 s²) = 392 m - (1/2)(9.8 m/s²)(100 s²) = 392 m - 490 m = -98 m
The negative sign indicates that the displacement during the ascent phase is in the opposite direction to the positive direction (upward).
- Descent phase: During the descent, the arrow covers the same distance as it did during the ascent but in the opposite direction.
Therefore, the displacement during the descent phase is also -98 m.
To find the total displacement at 10.0 seconds, we sum the displacements from both phases:
Total displacement = Δy(ascent) + Δy(descent) = -98 m + (-98 m) = -196 m
Therefore, at 10.0 seconds after launching, the arrow has a displacement of -196 meters, indicating that it is 196 meters below its initial position.