Question 30 ( 6 points) A motorcycle stunt driver accelerates horizontally off the edge of a 50.0 -meter-high cliff. To land on level ground 90.0 meters away from the base of the cliff, what minimum speed must the motorcycle have when it leaves the cliff? Ignore air resistance.

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UpStudy AI · Accepted Answer

To solve this problem, we need to determine the minimum speed at which the motorcycle must leave the cliff to land 90.0 meters away from the base of the cliff. We can break this problem into two parts: the vertical motion and the horizontal motion.

### Step 1: Analyze the vertical motion

The motorcycle falls from a height of 50.0 meters. We can use the following kinematic equation to find the time $$ t $$ it takes to fall this distance:

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

Where:
- $$ h = 50.0 $$ m (height of the cliff)
- $$ g = 9.81 $$ m/s² (acceleration due to gravity)

Rearranging the equation to solve for $$ t $$:

$$
t = \sqrt{\frac{2h}{g}}
$$

### Step 2: Calculate the time of fall

Now we can substitute the known values into the equation:

$$
t = \sqrt{\frac{2 \times 50.0}{9.81}}
$$

### Step 3: Analyze the horizontal motion

The horizontal distance $$ d $$ the motorcycle travels is given as 90.0 meters. The horizontal motion can be described by the equation:

$$
d = v_x \cdot t
$$

Where:
- $$ d = 90.0 $$ m (horizontal distance)
- $$ v_x $$ is the horizontal speed (which we need to find)

Rearranging the equation to solve for $$ v_x $$:

$$
v_x = \frac{d}{t}
$$

### Step 4: Calculate $$ v_x $$

Now we will calculate $$ t $$ first and then use it to find $$ v_x $$. Let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\frac{2\times 50}{9.81}}$$
- step1: Multiply the numbers:
  $$\sqrt{\frac{100}{9.81}}$$
- step2: Divide the numbers:
  $$\sqrt{\frac{10000}{981}}$$
- step3: Use the properties of radicals:
  $$\frac{\sqrt{10000}}{\sqrt{981}}$$
- step4: Simplify the expression:
  $$\frac{100}{3\sqrt{109}}$$
- step5: Multiply by the Conjugate:
  $$\frac{100\sqrt{109}}{3\sqrt{109}\times \sqrt{109}}$$
- step6: Multiply the numbers:
  $$\frac{100\sqrt{109}}{327}$$
Calculate or simplify the expression $$ 90.0/(3.192754) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{90}{3.192754}$$
- step1: Convert the expressions:
  $$\frac{90}{\frac{1596377}{500000}}$$
- step2: Multiply by the reciprocal:
  $$90\times \frac{500000}{1596377}$$
- step3: Multiply:
  $$\frac{90\times 500000}{1596377}$$
- step4: Multiply:
  $$\frac{45000000}{1596377}$$
### Step 1: Calculate the time of fall

Using the formula for the time of fall:

$$
t = \sqrt{\frac{2 \times 50.0}{9.81}} \approx 3.19 \text{ seconds}
$$

### Step 2: Calculate the minimum horizontal speed

Now, using the time calculated to find the horizontal speed:

$$
v_x = \frac{90.0}{3.19} \approx 28.19 \text{ m/s}
$$

### Final Answer

The minimum speed at which the motorcycle must leave the cliff to land 90.0 meters away is approximately **28.19 m/s**.
The motorcycle must leave the cliff at a minimum speed of approximately 28.19 m/s to land 90.0 meters away.

Question 30 ( 6 points) A motorcycle stunt driver accelerates horizontally off the edge of a 50.0 -meter-high cliff. To land on level ground 90.0 meters away from the base of the cliff, what minimum speed must the motorcycle have when it leaves the cliff? Ignore air resistance.

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