A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.40 s later.
What is the rocket's acceleration?
Figure the vertical velocity and height the bolt when it breaks off, in terms of a.
Then, using that hi, and vi, you can figure a from that equation.
hf=hi+Vi*time-1/2 g time^2 you know Vi, Hi in terms of a, and hf=0
There is a little algebra in this.
To find the rocket's acceleration, we can use the kinematic equation that relates distance, initial velocity, time, and acceleration. The equation is:
𝑑 = 𝑣₀𝑡 + ½𝑎𝑡²
Where:
- 𝑑 is the distance traveled
- 𝑣₀ is the initial velocity
- 𝑡 is the time
- 𝑎 is the acceleration
We are given that the bolt falls off the side of the rocket 4 seconds after liftoff and hits the ground 6.40 seconds later. This means the total time elapsed for the bolt to hit the ground is 6.40 seconds - 4 seconds = 2.40 seconds.
Let's rearrange the equation to solve for acceleration:
𝑎 = (𝑑 - 𝑣₀𝑡) / ½𝑡²
We need to find the distance 𝑑 and the initial velocity 𝑣₀.
Since the bolt falls freely, its initial velocity is 0 m/s. Therefore, 𝑣₀ = 0 m/s.
Now let's find the distance traveled by the bolt before hitting the ground. We can use the equation:
𝑑 = 𝑣₀𝑡 + ½𝑎𝑡²
Substituting the known values, 𝑣₀ = 0 m/s and 𝑡 = 2.40 s, we have:
𝑑 = 0 + ½𝑎(2.40)²
𝑑 = ½𝑎(5.76)
𝑑 = 2.88𝑎
Now we substitute the value of 𝑑 into the equation to solve for acceleration:
𝑎 = (2.88𝑎 - 0) / ½(2.40)²
𝑎 = 2.88𝑎 / ½(5.76)
𝑎 = 2.88𝑎 / (2.88)
𝑎 = 1
Therefore, the rocket's acceleration is 1 m/s².