To find the launch velocity of the projectile, we need to analyze the motion using two components: horizontal and vertical. Here's the breakdown of the problem:

1. Given Information:

• Height of the building ($h$): 85 m
• Horizontal distance covered ($d_x$): 99 m
• Time of flight ($t$): 6.0 s
• Gravity ($g$): $9.8 \, \text{m/s}^2$

We need to find the initial velocity ($v_0$).

2. Horizontal Motion:

The horizontal velocity ($v_{0x}$) is constant throughout the motion, since there is no horizontal acceleration.

The formula for horizontal displacement is: $d_x = v_{0x} \cdot t$ Thus, $v_{0x} = \frac{d_x}{t} = \frac{99}{6.0} = 16.5 \, \text{m/s}$

3. Vertical Motion:

The vertical motion is influenced by gravity. The formula for vertical displacement is: $y = v_{0y} \cdot t - \frac{1}{2} g t^2$ Where:

• $y$ is the vertical displacement, which is $-85 \, \text{m}$ (since the projectile ends up 85 m below its initial position)
• $v_{0y}$ is the initial vertical velocity (unknown)
• $g = 9.8 \, \text{m/s}^2$
• $t = 6.0 \, \text{s}$

Plugging in the values: $-85 = v_{0y} \cdot 6.0 - \frac{1}{2} \cdot 9.8 \cdot (6.0)^2$ $-85 = 6.0 v_{0y} - \frac{1}{2} \cdot 9.8 \cdot 36$ $-85 = 6.0 v_{0y} - 176.4$ $6.0 v_{0y} = 91.4$ $v_{0y} = \frac{91.4}{6.0} = 15.23 \, \text{m/s}$

4. Total Launch Velocity:

The total launch velocity $v_0$ can be found using the Pythagorean theorem, since the initial velocity has both horizontal and vertical components: $v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$ $v_0 = \sqrt{(16.5)^2 + (15.23)^2}$ $v_0 = \sqrt{272.25 + 231.93} = \sqrt{504.18}$ $v_0 \approx 22.45 \, \text{m/s}$

Final Answer:

The launch velocity of the projectile is approximately 22.45 m/s.

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Do you need more details on any step or have any questions?

Related Questions:

1. What is the angle of launch given the initial velocity components?
2. How would the launch velocity change if the time of flight was different?
3. What would the horizontal velocity be if the projectile landed 120 m from the base?
4. How long would the projectile be in the air if the height of the building was 100 m instead?
5. How does the effect of air resistance alter the projectile's motion in real-life scenarios?

Tip:

Always break down projectile motion into horizontal and vertical components for simpler calculations!