5. an arrow is shot from a bow straight upwards at an initial speed of 104 m/s. how high in the air does the…
5. an arrow is shot from a bow straight upwards at an initial speed of 104 m/s. how high in the air does the arrow fly before it comes back down?
Answer
Explanation:
Step1: Identify the formula
Use the kinematic equation (v^{2}=v_{0}^{2}+2ah). At the maximum height, (v = 0), (v_{0}=104\ m/s), and (a=-g=- 9.8\ m/s^{2}).
Step2: Rearrange the formula for (h)
From (v^{2}=v_{0}^{2}+2ah), we get (h=\frac{v^{2}-v_{0}^{2}}{2a}).
Step3: Substitute the values
Substitute (v = 0), (v_{0}=104\ m/s), and (a=-9.8\ m/s^{2}) into the formula: (h=\frac{0-(104)^{2}}{2\times(-9.8)}=\frac{- 10816}{-19.6})
Step4: Calculate the value of (h)
(h = 552\ m)
Answer:
The arrow flies (552\ m) high in the air before it comes back down.