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 horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22..(I got 40 m/s)
B)If the ramp is now tilted upward, so that "takeoff angle" is 7.0∘ above the horizontal, what is the new minimum speed? (I cant figure out)
Okay, I was having problems with this and tried twisted the numbers around, not sure if right but it could be.
A) Vi=39.8 m/s
B)Y=Yi+VyiT+(1/2)a(t^2)
use X=Xo+VxiT
to make T=x/Vxi
This way you use substitution
Y=Yi+Vyi(X/Vxi)+(1/2)a(X/Vxi)^2
When you look at the vectors you can relate the angle and the two velocity vectors using tan(7)=(Vyi/Vxi)
Substitute that back in and you get
Y=Yi+tan(7)X+(1/2)a(X/Vxi)^2
Y=final Y so Y=-1.5m
X=final X so 22m
a=-9.8m/s^2
Using all of this you can find Vxi and from there get the other component as well as the answer
I find that the answer is...
Vxi=23.76m/s
Vyi-2.92m/s
Vinitial=23.94m/s
To determine the minimum speed required for the car to clear the ramp and jump over the parked cars, we can use the principles of projectile motion.
A) Without any tilt in the ramp (horizontal ramp):
We need to find the minimum speed required to clear the horizontal distance of 22 m and the vertical height of 1.5 m.
1. To clear the horizontal distance, we can use the equation: d = v * t, where d is the horizontal distance, v is the horizontal velocity, and t is the time of flight.
2. From the equation above, we can find the time of flight using: t = d / v.
3. To clear the vertical distance, we can use the equation: h = 0.5 * g * t^2, where h is the vertical height and g is the acceleration due to gravity (approximately 9.8 m/s^2).
4. Rearranging the equation from step 3, we can solve for t: t = sqrt(2 * h / g).
5. Equate the expressions for t from steps 2 and 4: d / v = sqrt(2 * h / g).
6. Rearrange the equation from step 5 to solve for v: v = d / sqrt(2 * h / g).
Substituting the given values:
d = 22 m,
h = 1.5 m,
g = 9.8 m/s^2.
v = 22 / sqrt(2 * 1.5 / 9.8) = 40 m/s.
Therefore, the minimum speed required for the car to clear the ramp without any tilt is 40 m/s.
B) With a tilt of 7.0 degrees above the horizontal:
In this case, the vertical component of the car's velocity will be affected.
1. We can use the same formula as in step 6, but now we need to consider the vertical component of velocity (v_y) as well: v = sqrt(v_x^2 + v_y^2), where v_x represents the horizontal component of velocity.
2. The horizontal component of velocity can be found using: v_x = v * cos(theta), where theta is the takeoff angle (7.0 degrees in this case).
3. The vertical component of velocity can be found using: v_y = v * sin(theta).
4. Using the equation in step 1, we can substitute the expressions for v_x and v_y: v = sqrt((v * cos(theta))^2 + (v * sin(theta))^2).
5. Rearrange the equation from step 4 to solve for v: v = sqrt((v^2 * cos^2(theta)) + (v^2 * sin^2(theta))).
6. Simplify the equation from step 5: v = v * sqrt(cos^2(theta) + sin^2(theta)).
7. Since cos^2(theta) + sin^2(theta) = 1, the equation from step 6 becomes: v = v * sqrt(1) = v.
Therefore, the minimum speed required to clear the ramp with a tilt of 7.0 degrees above the horizontal will be the same as before, which is 40 m/s.
To find the minimum speed required for the car to jump over the cars, we need to consider the physics principles involved in projectile motion. Let's break down the problem step by step.
A) The first part of the question asks for the minimum speed the car must have to clear 8 cars that are parked side by side below a horizontal ramp, with a height of 1.5 meters above the cars and a horizontal distance of 22 meters.
To solve this, we'll use the equations of projectile motion. The key idea is that the horizontal and vertical motions are independent of each other.
1. Start by analyzing the vertical motion. The car must clear a vertical distance of 1.5 meters. We'll use this to find the initial vertical velocity (Vyi).
Using the equation:
Vf^2 = Vi^2 + 2aΔy
Where Vf is the final velocity (which is zero at the top of the jump), Vi is the initial velocity, a is the acceleration (which is -9.8 m/s^2 due to gravity), and Δy is the vertical displacement.
Plug in the values:
0 = Vi^2 + 2(-9.8)(1.5)
Solving this equation will give us the initial vertical velocity (Vyi).
2. Next, we'll analyze the horizontal motion. The car needs to travel a horizontal distance of 22 meters. We can use the equation for horizontal velocity and time:
Vxi = Δx / t
Where Vxi is the initial horizontal velocity, Δx is the horizontal distance, and t is the time of flight.
3. Now, we can find the total initial speed (Vi) using the Pythagorean theorem:
Vi = √(Vxi^2 + Vyi^2)
Substitute the values of Vxi and Vyi from the previous steps into this equation to get the minimum speed required for the car to clear the 8 cars.
B) In the second part of the question, the ramp is tilted upward, with a "takeoff angle" of 7.0 degrees above the horizontal. We need to find the new minimum speed required for the car to clear the cars.
To do this, we follow the same steps as in Part A, but with a few modifications. The main change is that the takeoff angle affects the horizontal and vertical velocities of the car.
1. Determine the new vertical velocity (Vyi) by multiplying the initial speed (Vi) by the sine of the takeoff angle.
Vyi = Vi * sin(θ)
Where θ is the takeoff angle (7.0 degrees).
2. Determine the new horizontal velocity (Vxi) by multiplying the initial speed (Vi) by the cosine of the takeoff angle.
Vxi = Vi * cos(θ)
3. Use the equations explained in Part A to calculate the time of flight, initial vertical velocity, and minimum speed required for the car to clear the 8 cars.
By following these steps, you will be able to find the minimum speed required for the car to successfully clear the cars, both in the case of a horizontal ramp and a ramp tilted upwards. Plug in the appropriate values to get the final answers.
a= 37m/s
b= 20m/s