A stunt driver wants to make his car jump over 8 cars parked side by side below

Question

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp. It is 1.5m above the cars, and it must travel 22m to land safely.
a)calculate the minimum speed it must travel over the horizontal ramp to barely clear the cars.
b)if the ramp is now tilted upwards at 7 degrees above the horizontal, what is the new minimum speed he must drive off the ramp to barely clear the cars safely.

a)for a I found the instantaneous velocity of the car before it hit to ground to be -5.42m/s, I used this to find the time it was falling to be 0.55s, and then used this to find the velocity needed to be 40m/s.

b) I have tried many ways to find b and all have been wrong. I need much guidance here. any help would be appreciated

Explanation / Answer

(a) Vertical height, h = 1.5 m Horizontal distance, X = 22 m time taken, t = v ( 2h / g ) = v ( 3/ 9.8) = 0.553 s X = U t Minimum Horizontal velocity, U = X / t = 39.78 m/s (b) -h = U sin ? - ( 1/2) g t2 - 1.5 = U Sin 7 o - 4.9 t2 .......................(1) X = U cos ? (t) 22 = U cos 7o(t) .........................(2) Solve the above two equations to get 'U'

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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