Question

A “Stomp rocket” is a toy rocket that is propelled by stomping on a plastic bottle attached to the rocket stand by a rubber hose. One really good “stomp” is capable of propelling the rocket with an initial velocity of = 22 m/s. a. (3 points) Find an expression for the time the rocket is in the air, assuming it is launched over level ground. b. (3 points) Derive the expression for the horizontal distance (the range) traveled by the rocket. c. (3 points) Using the formula derived in part (b), find the maximum range of the rocket in meters.

Answer 1

a. To find an expression for the time the rocket is in the air, we can use the following kinematic equation for vertical motion:

\[d = v_i t + \frac{1}{2} a t^2\]

Where:

- \(d\) is the vertical displacement (which is zero since the rocket lands at the same level it was launched),

- \(v_i\) is the initial vertical velocity (22 m/s, upward),

- \(a\) is the acceleration due to gravity (-9.8 m/s², acting downward), and

- \(t\) is the time we want to find.

Since the rocket is launched vertically, it reaches its maximum height before falling back down. At the maximum height, the vertical velocity becomes zero. Therefore:

\[0 = 22 m/s - 9.8 m/s² t\]

Solving for \(t\):

\[9.8 m/s² t = 22 m/s\]

\[t = \frac{22 m/s}{9.8 m/s²}\]

Now, calculate \(t\):

\[t \approx 2.24\ \text{seconds}\]

So, the time the rocket is in the air is approximately 2.24 seconds.

b. To derive the expression for the horizontal distance (the range) traveled by the rocket, we can use the following formula:

\[R = v_x t\]

Where:

- \(R\) is the horizontal range (distance traveled),

- \(v_x\) is the horizontal component of the initial velocity (since there is no horizontal acceleration, \(v_x\) remains constant), and

- \(t\) is the time calculated in part (a), which is approximately 2.24 seconds.

The horizontal component of the initial velocity is the same as the initial velocity (22 m/s) because there is no horizontal acceleration:

`\[v_x = 22\ \text{m/s}\]`

Now, plug in the values to find the range:

`\[R = (22\ \text{m/s}) \cdot (2.24\ \text{s})\]`

`\[R \approx 49.28\ \text{meters}\]`

So, the horizontal distance (range) traveled by the rocket is approximately 49.28 meters.

c. Using the formula derived in part (b), we can find the maximum range of the rocket by plugging in the value of \(v_x\) (horizontal component of the initial velocity, which is the same as the initial velocity) and \(t\) (the time calculated in part a):

`\[R_{\text{max}} = (22\ \text{m/s}) \cdot (2.24\ \text{s})\]`

`\[R_{\text{max}} \approx 49.28\ \text{meters}\]`

So, the maximum range of the rocket is approximately 49.28 meters.