#EXAMPLE: A stunt man drives a car at a speed of 20m/s off a 31m high cliff. The

Question

#EXAMPLE: A stunt man drives a car at a speed of 20m/s off a 31m high cliff. The road leading to the cliff is inclined upward at an angle of 20 degree.
a)How far from the base of the cliff does the car land?
#solution: x=62m
b)What is the car's impact speed?
#solution: v=32m/s

*QUESTION: A stunt man drives a car at a speed of 20m/s off a 29m high cliff. The road leading to the cliff is inclined upward at an angle of 20 degree.
a)How far from the base of the cliff does the car land?
*Solution:??????
b)What is the car's impact speed?
*Solution: ????????

NOTE:
1)plz answer the question. I want to make it easier for you so that I post an example with its answer. you notice its the same question but different numbers, so plz answer the question correctly.
2) plz indicate the question number when you answer, for instance:
a).........
b).........

THAX

Explanation / Answer

a. D= S+Y. = 59.74m s= u cosx t t= 2usinx/g s= u^2 sin2x /g = 400 X sin40 /9.8 = 23.99 . this is the distance at the level 29m . now from this level, the horizontal movement(let it be Y) is due to the horizontal component of the velocity, which is ucos x and time of travel is obtained by equation, 29 = usinx + 1/2 gt^2 . x= 20 degrees. t= 1.88 s now apply ucosx X t = 35.75 and add this distance to ie... b. 31.08 m/s the impact velocity is calculated by , v= ucosx i + (usinx + gt) where t is the time calculated in the above part. then , find out the magnitde of the vector.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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