A new amusement park ride consists of a combination of physical principles. The

Question

A new amusement park ride consists of a combination of physical principles. The ride is a water flume. where you ride in an open car on a layer of water. The ride takes the 2.0 m long car to the top of a hill where you stop momentarily before plunging down a steep incline. As you go up the second hill, the ramp ends at a point 17 m above the ground at an angle of 50degree to the horizontal. The car then flies through the air and lands in a pool of water. The car's mass is 100 kg and the ride restricts passengers to be at least 50 kg and no more than 120 kg. Friction on the slide does a total of 5000 J of work. How far from the end of the ramp should the pool be placed and what is the minimum width of the pool considering the weight restriction? If there is a concession stand between the ride and the pool, how high above the ground can a flag on top of the stand be in order for the car to just clear it?

Explanation / Answer

Let us first calculate the velocity at height = 17m

Use equation,

KEf = PEi - Wfric

1/2Mvf^2 = Mgh - Wfric

To get minimum vf M should be maximum

Thus M = mass of car + max mass of passengers = 100kg + 120kg= 220kg

Plugging values,

½*220*vf^2 = 220*9.8*35 – 5000    => vf= 25.31m/s

vfx=vf*cos = 25.31cos50 = 16.27m/s

vfy=vf*sin = 25.31sin50 = 19.39 m/s

Now time required to land in the pool from the height 17m = t

h=viy*t – 1/2gt^2

-17.0 = 19.39*t – ½*9.8*t^2   => t= 4.7s

We have to find horizontal distance X,

X= vfx*t = 16.27*4.7 = 76.45m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.