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A missile is shot upward with a velocity of 189 m/s from the top of a 20 m cliff. How much
time will it take for the missile to fall on the ground?

To solve this problem, we'll need to apply the equations of motion under gravity. The missile is launched upwards with an initial velocity and will eventually fall back to the ground. We'll solve this in two steps: first, find the time it takes for the missile to reach its peak height, and second, find the time it takes to fall from the peak height to the ground.

### Step 1: Time to reach the maximum height
At the maximum height, the missile's velocity becomes zero. We can use the following equation:

\[
v = u - g t_1
\]

Where:
- \( v = 0 \) m/s (final velocity at the peak),
- \( u = 189 \) m/s (initial velocity),
- \( g = 9.8 \) m/s² (acceleration due to gravity),
- \( t_1 \) is the time taken to reach maximum height.

Solving for \( t_1 \):

\[
0 = 189 - 9.8 \times t_1
\]

\[
t_1 = \frac{189}{9.8} \approx 19.29 \text{ seconds}
\]

### Step 2: Maximum height
We can find the maximum height using the following equation:

\[
h = u t_1 - \frac{1}{2} g t_1^2
\]

Substituting the values:

\[
h = 189 \times 19.29 - \frac{1}{2} \times 9.8 \times (19.29)^2
\]

\[
h \approx 1823.21 \text{ meters}
\]

### Step 3: Total height above the ground
The missile was launched from a height of 20 meters, so the total height above the ground at the peak is:

\[
H_{\text{total}} = 1823.21 + 20 = 1843.21 \text{ meters}
\]

### Step 4: Time to fall from the peak to the ground
The time to fall can be found using the equation:

\[
H_{\text{total}} = \frac{1}{2} g t_2^2
\]

Where \( t_2 \) is the time taken to fall from the maximum height to the ground. Solving for \( t_2 \):

\[
1843.21 = \frac{1}{2} \times 9.8 \times t_2^2
\]

\[
t_2^2 = \frac{1843.21 \times 2}{9.8} \approx 376.16
\]

\[
t_2 \approx 19.39 \text{ seconds}
\]

### Step 5: Total time to hit the ground
The total time \( T \) for the missile to reach the ground is the sum of \( t_1 \) and \( t_2 \):

\[
T = t_1 + t_2 = 19.29 + 19.39 \approx 38.68 \text{ seconds}
\]

Thus, it will take approximately **38.68 seconds** for the missile to fall to the ground.

Would you like any further details or have any questions?

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Here are 5 related questions to consider:

1. How would the time change if the missile was launched from a greater height?
2. What would be the time of flight if the missile was launched at an angle?
3. How would air resistance affect the time it takes for the missile to hit the ground?
4. How can we calculate the velocity of the missile just before it hits the ground?
5. What if the missile was shot downward with the same initial speed; how would the time change?

**Tip:** When solving motion problems, always break down the motion into distinct phases (upward, peak, and downward) and apply the appropriate equations to each phase.

A missile is shot upward with a velocity of 189 m/s from the top of a 20 m cliff. How much time will it take for the missile to fall on the ground?

Solve a real-world kinematics problem: Determine the total time for a missile shot upwards at 189 m/s from a 20 m cliff to hit the ground. Learn step-by-step, using equations of motion under gravity. Ideal for high school physics students in Grades 10-12.

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