In a game a stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top to land on ground level below 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. [Hint : Split up into vertical and horizontal motion] [Solution 28.2 m/s]

To solve this problem, we can use the principles of projectile motion by splitting the motion into vertical and horizontal components.

### Given Data:
- Height of the cliff, \( h = 50.0 \, \text{m} \)
- Horizontal distance from the cliff to the landing point, \( d = 90.0 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The task is to find the initial horizontal velocity (\( v_{x0} \)) needed for the motorcycle to travel 90.0 m horizontally while falling from a height of 50.0 m.

### Solution:

#### Step 1: Calculate the time to fall vertically 50.0 m

Since the motorcycle is moving horizontally off the cliff, the initial vertical velocity (\( v_{y0} \)) is 0. The time \( t \) to fall can be found using the equation for vertical motion:

\[
h = \frac{1}{2} g t^2
\]

Rearrange to solve for \( t \):

\[
t = \sqrt{\frac{2h}{g}}
\]

Substitute the values:

\[
t = \sqrt{\frac{2 \times 50.0}{9.8}} = \sqrt{\frac{100}{9.8}} = \sqrt{10.204} \approx 3.2 \, \text{s}
\]

So, the time taken to fall to the ground is approximately \( t = 3.2 \, \text{s} \).

#### Step 2: Calculate the required horizontal velocity

Now, we know the motorcycle needs to travel 90.0 m horizontally in 3.2 s. The horizontal velocity \( v_{x0} \) can be found using:

\[
v_{x0} = \frac{d}{t}
\]

Substitute the values:

\[
v_{x0} = \frac{90.0}{3.2} \approx 28.2 \, \text{m/s}
\]

### Answer:
The motorcycle must leave the cliff with a horizontal speed of approximately **28.2 m/s** to land 90.0 m away from the base of the cliff.

Would you like more details on any of the steps, or have any further questions?

Here are some related questions:
1. How would the solution change if there were air resistance?
2. What if the cliff height were different, say 100 m?
3. How can we find the impact velocity of the motorcycle upon hitting the ground?
4. How does the launch angle affect the landing distance in projectile motion?
5. What is the effect of gravity on horizontal velocity in projectile motion?

**Tip:** In projectile motion problems, always separate the motion into horizontal and vertical components and use relevant kinematic equations for each component.

In a game a stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top to land on ground level below 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

Learn how to calculate the horizontal velocity needed for a motorcycle to jump off a 50-meter high cliff and land 90 meters away. This solution applies principles of projectile motion, splitting the motion into horizontal and vertical components. Suitable for high school students studying physics.

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