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]
Use the following:

x = x_0 + v_0 * t + ((a * t^2) / 2)

v = v_0 + a * t

To solve this problem, we need to split the motion into two components: vertical and horizontal. Since the motion is horizontal at the beginning, there is an initial horizontal velocity, but no initial vertical velocity.

### Given Information
- **Height of the cliff (vertical distance, \( y \))**: \( 50.0 \, \text{m} \)
- **Horizontal distance to reach the cameras (horizontal distance, \( x \))**: \( 90.0 \, \text{m} \)
- **Gravitational acceleration \( g \)** (since air resistance is ignored): \( 9.8 \, \text{m/s}^2 \)
- **Required final velocity for horizontal direction \( v_0 \):** Unknown (we need to find this)

### Step 1: Analyze Vertical Motion
Since the motorcycle is initially moving horizontally, there is no initial vertical velocity (\( v_{0y} = 0 \)).

Using the equation for vertical displacement:
\[
y = y_0 + v_{0y} \cdot t + \frac{1}{2} g t^2
\]

Since the cliff is 50.0 m high, we have:
\[
50.0 = 0 + 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2
\]

This simplifies to:
\[
50.0 = 4.9 t^2
\]

Solving for \( t \):
\[
t^2 = \frac{50.0}{4.9} \approx 10.2
\]
\[
t \approx \sqrt{10.2} \approx 3.2 \, \text{s}
\]

### Step 2: Analyze Horizontal Motion
In horizontal motion, there is no acceleration (ignoring air resistance), so the velocity remains constant.

The equation for horizontal displacement is:
\[
x = v_{0x} \cdot t
\]

We know \( x = 90.0 \, \text{m} \) and \( t = 3.2 \, \text{s} \), so:
\[
90.0 = v_{0x} \cdot 3.2
\]

Solving for \( v_{0x} \):
\[
v_{0x} = \frac{90.0}{3.2} \approx 28.2 \, \text{m/s}
\]

### Answer
The motorcycle must leave the cliff top at a speed of approximately **28.2 m/s** horizontally to land 90.0 m from the base of the cliff.

Would you like further details on this problem, or do you have any other questions?

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Here are 5 related questions for practice:
1. What would be the horizontal speed required if the cliff height were 100 m?
2. How long would it take for the motorcycle to hit the ground if the horizontal distance was only 50 m?
3. What would happen to the time of fall if air resistance was considered?
4. How would you calculate the final velocity of the motorcycle upon impact with the ground?
5. What is the effect of increasing the cliff height on the required horizontal velocity?

**Tip:** Always separate horizontal and vertical components in projectile motion problems to simplify your calculations.

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]

Learn how to calculate the speed a motorcycle must leave a 50.0m high cliff to land 90.0m away. This problem covers projectile motion, splitting horizontal and vertical components of motion, and uses kinematic equations. Suitable for high school students studying physics.

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