You shoot an arrow into the air. Two seconds later (2.00 {\rm s}) the arrow has gone straight upward to a height of 32.0 m above its launch point.
What is the initial velocity and how long did it take for the arrow to reach 16.0 m above the launch point?
This must be a repeat post; I recall answering it yesterday
To find the initial velocity and the time it takes for the arrow to reach a certain height, we can use the laws of motion under constant acceleration.
Let's start by finding the initial velocity. We can use the formula for displacement in vertical motion:
Δy = v₀t + (1/2)at²
where Δy is the displacement, v₀ is the initial velocity, t is the time, and a is the acceleration (in this case, due to gravity).
We are given that the arrow reaches a height of 32.0 m in 2.00 s. Plugging these values into the formula:
32.0 m = v₀(2.00 s) + (1/2)(9.8 m/s²)(2.00 s)²
Simplifying the equation:
32.0 m = 2.00v₀ + (1/2)(9.8 m/s²)(4.00 s²)
32.0 m = 2.00v₀ + 19.6 m/s²(4.00 s²)
32.0 m = 2.00v₀ + 78.4 m
-78.4 m = 2.00v₀
v₀ = -39.2 m/s
Therefore, the initial velocity of the arrow is -39.2 m/s. Note that the negative sign indicates that the arrow was shot upwards.
Now, let's find the time it takes for the arrow to reach 16.0 m above the launch point. We'll again use the formula for displacement, but this time the displacement is 16.0 m:
16.0 m = v₀t + (1/2)(9.8 m/s²)t²
Since we've already found the value of v₀, we can substitute it:
16.0 m = (-39.2 m/s)t + (1/2)(9.8 m/s²)t²
Rearranging the equation:
(1/2)(9.8 m/s²)t² - (-39.2 m/s)t + 16.0 m = 0
Now we have a quadratic equation in the form of at² + bt + c = 0. We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values:
t = [-(39.2 m/s) ± √((39.2 m/s)² - 4(1/2)(9.8 m/s²)(16.0 m))] / [2(1/2)(9.8 m/s²)]
Simplifying the equation:
t = [-(39.2 m/s) ± √(1536.64 m²/s² + 3136 m²)] / (2(4.9 m/s²))
t = [-(39.2 m/s) ± √(4672.64 m²)] / (9.8 m/s²)
t = [-(39.2 m/s) ± 68.32 m] / (9.8 m/s²)
We have two possible solutions for t:
t₁ = [-(39.2 m/s) + 68.32 m] / (9.8 m/s²) ≈ 3.21 s
t₂ = [-(39.2 m/s) - 68.32 m] / (9.8 m/s²) ≈ -11.80 s
Since time cannot be negative in this context, we discard the negative solution. Hence, the time it takes for the arrow to reach 16.0 m above the launch point is approximately 3.21 s.