Heres a tough physics question. can you answer it?

Question

A daredevil jumps a canyon 13 m wide. To do so, he drives a car up a 14° incline.
(a) What minimum speed must he achieve to clear the canyon?
m/s
(b) If the daredevil jumps at this minimum speed, what will his speed be when he reaches the other side?
m/s
good luck!!!

Answer 1

I can help you with that.

Part A:
Some things to notice: Vx will stay constant... Vy will change, due to acceleration due to gravity. g=NEGATIVE 9.81m/s^2, if Vy is initially positive.

Also, I'm going to assume that the place he jumps from is the same height as the place he lands... since it doesn't say so in your question.

First, you need to come up with equations for Vx and Vy:

Vx=Vcos(14)
Vy=Vsin(14)

Ok, now that we've defined those, let's go on to finding the equation for the TIME it takes him to get from one side of the canyon to the other... this may seem like a wierd place to start at, but it will make sense later...

Sx=Vx*t
13m=Vcos(14)*t
t=13m/(Vcos(14))

Now, we know that the total vertical distance the car travels is zero... so we use gthe following equation:

Sy=Vyt+(1/2)gt^2
0=Vsin(14)*t+(1/2)(9.81m/s^2)(t^2)

Now, if you substitute t=13m/(Vcos(14)) into this equation, you get everything in terms of V.

Solve it, using the quadratic equation, and you have your answer for Part A. (I'll leave that part to you.) Note that if the two sides of the canyon are not the same height, just change to Sy in the last equation into something other than zero.

Part B is actually really easy. We know that the Vx will not change... it never does in this kind of motion. So, you only have to find Vy. But that's also very easy. Since the overall change in y is zero, the Vy will be the same at the start of the jump, and at the end of the jump. Since Vx and Vy are the same as the start, V must be the same. Therefore, the answer to part B is the same as part A. (unless there is a differences in the height of the two sides of the canyon...)

I hope this helps you understand these kind of questions.

Answer 2

x= xo + vt + 1/2 at^2,
You want the x component of the Velocity.

V x component =cos 14 * Vo
substitute into the first equation i wrote...initial starting point is 0 and there is no horizontal acceleration.

13 = 0 + cos14 (vo) +0
V0 = 13/cos(14)
You do that math.

X motion is constant because unlike the Y motion, which is affected by gravity, there is no acceleration messing with X motion. Assuming the opposite side is flat, he will have the same speed as the X motion component, cos 14 (Vo)

Answer 3

mr, 13metres wide how longs the jump? cause if its only 13 metres long you want to be going like a mozzie after fresh blood at a speed no slower then 140 mls an hour witth a wind velosity of no more then 2nts and if you kiss your butt in a complex yoga move you might make it
im no physicist but i am a daredevil,.
please dont try this at home kids i am a certified idiot.

Answer 4

haha no clue who would do this for fun??

Answer 5

37mph

Answer 6

ooh..... fun with parabolas!