A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.(a)With what minimum speed must he drive off the horiontal ramp?The vertical height of the ramp is 1.5m above the cars and the horizontal distance must be clear is 20m.(b)If the ramp now tilted upwards,so that "take off angle" is 7.0 above the horizontal,what is the new minimum speed?
To solve this problem, we can use the principle of conservation of energy and apply some basic kinematic equations.
(a) To find the minimum speed required to clear the 8 cars parked side by side, we need to consider the height and horizontal distance of the ramp.
1. First, let's calculate the gravitational potential energy at the highest point of the ramp. The potential energy is given by the formula:
Potential energy = mass * gravity * height
Since we need to reach a height of 1.5m above the cars, we can assume the mass of the car doesn't affect this calculation. The acceleration due to gravity is approximately 9.8 m/s².
Potential energy = mass * 9.8 * 1.5
2. Next, we need to calculate the kinetic energy required to clear the horizontal distance of 20m. The kinetic energy is given by the formula:
Kinetic energy = 0.5 * mass * velocity²
Since we are looking for the minimum speed, we assume the car is just at the edge of leaving the ramp.
3. Now, let's analyze the total mechanical energy at the highest point of the ramp. At the highest point, the total mechanical energy (the sum of kinetic and potential energies) is equal to the kinetic energy required to clear the horizontal distance.
Total mechanical energy = potential energy + kinetic energy
Since we're considering the minimum speed, the total mechanical energy will be at its minimum value, which is equal to the kinetic energy required to clear the distance.
4. Equating the total mechanical energy to the kinetic energy required, we can solve for the minimum speed (velocity):
0.5 * mass * velocity² = mass * 9.8 * 1.5
Simplifying the equation, we find:
velocity² = 2 * 9.8 * 1.5
velocity = √(2 * 9.8 * 1.5)
The minimum speed required to clear the 8 parked cars is the square root of twice the product of 9.8 and 1.5.
(b) Now, let's consider the ramp tilted upwards with a "take off angle" of 7.0° above the horizontal. We need to find the new minimum speed required using the same principle of conservation of energy.
1. First, we need to split the velocity into horizontal and vertical components. The horizontal component remains the same, but the vertical component needs to be calculated using the angle.
vertical velocity = horizontal velocity * tan(angle)
2. Now, let's calculate the new vertical displacement (height) based on the given angle and the initial vertical velocity at takeoff.
height = (vertical velocity²) / (2 * acceleration due to gravity)
3. Using the same calculation as before, we can find the minimum speed required to reach this new height and clear the horizontal distance:
velocity = √(2 * gravity * height)
The minimum speed required will be the square root of twice the product of gravity and the new height.
By applying these steps, you can solve both parts of the given problem.