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?

tanA = 1.5/20 = 0.075,

A = 4.3 Deg.

a. Vf^2 = Vo^2 + 2gd,
Vo^2 = Vf^2 - 2gd,
Vo^2 = 0 - 2*(-9.8)*1.5 = 29.7,
Vo(v) = 5.45m/s. = ver. component.
Vo = 5.45 / sin4.3 = 72.7m/s = Inital
velocity.

wat about b?

To solve this problem, we will use the principles of kinematics and the laws of motion. Let's break it down into two parts:

a) First, let's calculate the minimum speed the stunt driver needs to drive off the horizontal ramp to clear the 8 cars parked side by side.

Given data:
Vertical height of the ramp (h) = 1.5 m
Horizontal distance to clear (d) = 20 m

To find the minimum speed (v), we can use the equations of motion in the vertical direction.

The key is to consider the conservation of energy. At the topmost point of the jump, all of the car's initial potential energy (mgh, where m is the mass of the car and g is the acceleration due to gravity) is converted into kinetic energy (1/2 mv^2, where v is the velocity of the car). Therefore, we can equate these two energies:

mgh = (1/2) mv^2
gh = (1/2) v^2
v^2 = 2gh
v = sqrt(2gh)

Now plug in the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.5 m

v = sqrt(2 * 9.8 * 1.5)
v = sqrt(29.4)
v ≈ 5.42 m/s

Therefore, the stunt driver must have a minimum speed of approximately 5.42 m/s to clear the 8 cars parked side by side.

b) Now let's consider the case where the ramp is tilted upwards, so the "take off angle" is 7.0 degrees above the horizontal. We need to find the new minimum speed required.

Given data:
Take off angle (θ) = 7.0 degrees
Vertical height of the ramp (h) = 1.5 m
Horizontal distance to clear (d) = 20 m

To solve this, we need to consider the horizontal and vertical components of the velocity.

The horizontal component of the velocity (vx) will remain constant, as there is no horizontal force acting on the car. Therefore, vx = v (where v is the total velocity of the car).

The vertical component of the velocity (vy) can be calculated using the following equation:

vy = v * sin(θ)

Now, we can apply the equations of motion in the vertical direction. The initial velocity (uy = vy) is zero because the car starts from rest vertically. The final velocity (vfy) is also zero because the car reaches its maximum height at the topmost point.

Using the equation of motion:
vfy^2 = vuy^2 + 2aΔy

Since vfy = 0 and Δy = h, we have:
0 = (v * sin(θ))^2 + 2 * (-g) * h

Simplifying, we get:
0 = v^2 * sin^2(θ) - 2gh

Finally, solving for v:
v^2 = (2gh) / sin^2(θ)
v = sqrt((2gh) / sin^2(θ))

Now, plug in the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.5 m
θ = 7.0 degrees

v = sqrt((2 * 9.8 * 1.5) / sin^2(7.0))
v ≈ 5.45 m/s

Therefore, the new minimum speed required with the ramp tilted upwards is approximately 5.45 m/s.