To find the initial speed with which an arrow must be fired straight upward to reach a maximum height of 110 meters in 5.4 seconds, we can use the kinematic equations of motion. Specifically, we can use the equation that relates displacement, initial velocity, and acceleration:

$h = v_0 t - \frac{1}{2} g t^2$

Where:

• $h = 110 \, \text{m}$ is the maximum height,
• $t = 5.4 \, \text{seconds}$ is the time to reach that height,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $v_0$ is the initial speed we are solving for.

Now, rearrange the equation to solve for $v_0$:

$v_0 = \frac{h + \frac{1}{2} g t^2}{t}$

Substituting the known values:

$v_0 = \frac{110 + \frac{1}{2} \times 9.8 \times (5.4)^2}{5.4}$

Let's calculate that.The initial speed required to fire the arrow straight upward to reach a maximum height of 110 meters in 5.4 seconds is approximately $46.83 \, \text{m/s}$.

Would you like more details on this calculation or have any further questions?

Here are 5 related questions for practice:

1. What would be the time to reach a maximum height of 150 m with the same initial speed?
2. How long would the arrow take to fall back to the ground after reaching the peak?
3. What is the arrow's velocity when it returns to the ground?
4. How would air resistance affect the initial velocity needed to reach 110 m?
5. If the arrow was fired at an angle, how would that change the calculation?

Tip: When solving kinematic problems, always break the motion into vertical and horizontal components if there's an angle involved.