Projectile motion of car driving off a cliff

In summary, we can use the equation v = vi + at to find the horizontal and vertical velocities of a 1200kg car that drives off a cliff with a horizontal velocity of 30m/s and an upward vertical velocity of 15m/s. Neglecting air resistance, the horizontal velocity after 1s is 30m/s and the vertical velocity after 1s is 25m/s. To calculate the maximum height above the ramp the car reaches, we use the equation d = 1/2(vf + vi)t, which results in a maximum height of 11.25m. To determine whether the car will make the jump if the distance to the other side is 55m, we can check
  • #1
Alyssa Jesse
28
0
The question is -

A 1200kg car drives off a cliff with a horizontal velocity of 30m/s and an upward vertical velocity of 15m/s.
a) Neglecting air resistance, what is its horizontal and vertical velocity after 1s?
b) What is the maximum height above the ramp the car reaches?

I used - v = vi + at

so - vh = 30m/s + 0*1 = 30m/s
and - vv = 15m/s + 10*1 = 25m/s

However I am not sure if this is the right formula, and I can't see how to get from question a to question b. If someone could point me in the right direction in regards to formula it would be much appreciated. Thank you :)
 
Physics news on Phys.org
  • #2
Alyssa Jesse said:
The question is -

A 1200kg car drives off a cliff with a horizontal velocity of 30m/s and an upward vertical velocity of 15m/s.
a) Neglecting air resistance, what is its horizontal and vertical velocity after 1s?
b) What is the maximum height above the ramp the car reaches?

I used - v = vi + at

so - vh = 30m/s + 0*1 = 30m/s

This is correct.

and - vv = 15m/s + 10*1 = 25m/s

Why is this PLUS 10*1? What is the direction of acceleration relatively to the direction of initial vertical velocity?

However I am not sure if this is the right formula, and I can't see how to get from question a to question b. If someone could point me in the right direction in regards to formula it would be much appreciated. Thank you :)

What can you say about the vertical velocity when the maximum height is reached?
 
  • #3
Ahh, it should be -10*1 because gravity is causing it to accelerate in the opposite direction? So vv = 5m/s?
Vertical velocity when maximum height is reached would be 0m/s before beginning to fall..
 
  • #4
Alyssa Jesse said:
Ahh, it should be -10*1 because gravity is causing it to accelerate in the opposite direction? So vv = 5m/s?

Correct. Always mind the directions and the corresponding signs. They are important.

Vertical velocity when maximum height is reached would be 0m/s before beginning to fall..

Right. Now, does that allow you to compute when this will happen? And knowing when, can you compute the vertical distance?
 
  • #5
I'm not sure.. I can see there is something there in the car losing 10m/s in 1second but I'm not sure how to calculate how long it would take for it to get to 0m/s.
 
  • #6
would it be vi + vf/2? so (15m/s + 0m/s)/2 =7.5m?
 
  • #7
Alyssa Jesse said:
I'm not sure.. I can see there is something there in the car losing 10m/s in 1second but I'm not sure how to calculate how long it would take for it to get to 0m/s.

From your equation, ## V_v = V_{vi} - gt ##, at what ## t ## will ## V_v ## be zero?

Alyssa Jesse said:
would it be vi + vf/2? so (15m/s + 0m/s)/2 =7.5m?

No, this cannot be correct. Dimensionally this is velocity, while you need distance.
 
  • #8
I must of rearranged this formula wrong. I tried to solve for time by -

vv = vvi - (g*t)
vv - vvi = -(g*t)
(vv - vvi) / -g = -t
T = (vv - vvi)/g

so t = (0-15) / 10
so t= -1.5s

Which isn't a practical solution, as we don't work in negative time!
 
  • #9
Alyssa Jesse said:
I must of rearranged this formula wrong. I tried to solve for time by -

vv = vvi - (g*t)
vv - vvi = -(g*t)
(vv - vvi) / -g = -t

Here is your mistake. You had one minus on the right hand side. Then you moved it to the left hand side, AND retained on the right. That's not correct, because that essentially discards it completely, giving you the wrong sign in the end.
 
  • #10
Thank you!

t = (vv - vvi) / -g
t= (0-15)/-10
t= 1.5s

so,

d= 1/2 (vf + vi)*t
d= 1/2 (0 + 15)*1.5
d= 0.05m?
 
  • #11
How can (1/2) * (15) * (1.5) be equal to 0.05?
 
  • #12
Incorrect use of parentheses! Just testing you :P

d=11.25m

I think I need to go double check all my maths for the rest of my assignment now...
 
  • #13
Looking good now!
 
  • #14
The last part of this question is -

Calculate whether the car will make the the jump if (the distance to the other side) d=55m?

I don't know how to do this question without knowing the height of the cliff the car is on, and the height that it has to reach on the other side. Does it have something to do with the previous answer of it taking 1.5s to reach a height of 11.25m?
 
  • #15
I would guess - guess - that the initial and final points of the jump are at an equal height.
 
  • #16
The problem is I'm not told what the initial and final heights of the jump are, so how can I express this?
 
  • #17
As I said: I think they should be considered equal. In that case, does it really matter what their heights are?
 
  • #18
Hmm I understand the concept behind what you are saying, but I don't know how to show that using physics.
 
  • #19
What does "making the jump" mean?
 
  • #20
That the car will reach the other side without falling into the abyss?
 
  • #21
So how can you check that when the car reaches the other side, it has not fallen into the abyss?
 
  • #22
Well taking into consideration the previous problem where the car reaches 11.25m above the ramp, ie the initial starting point, and if the ramp is the same on the other side - that the car does not drop below 11.25m? That if the car drops below 11.25m then it would miss the ramp?
 
  • #23
11.25 m is the max height the car reaches ABOVE the cliff. It should not drop BELOW the cliff while it flies over the abyss.
 
  • #24
Ah sorry I wrote the last part wrong. If it drops lower than 11.25m below its highest point of 11.25m, then it would not make the jump?
 
  • #25
Well, that is definitely correct, but is it not simpler to say that "not making the jump" is when it drops below the initial height, i.e, zero?

Anyway, either if these two conditions is correct. So how can you check for those conditions?
 
  • #26
Yeah, you're right, it is simpler to think of it as initial height=zero!

Is this right?

d= vv+1/2at^2
d= 0+1/2(10)(1.5^2)
d= 11.25m

So it does make the jump?
 
  • #27
But the question asks, does it make the jump if d= 55m..I feel confused.
 
  • #28
Of course in 1.5 s the car will be at 11.25 m high - you computed that previously. But what does 1.5 s have to do with the question? The question asks about making a jump of 55 m, not 1.5 s!
 

1. What factors affect the projectile motion of a car driving off a cliff?

The main factors that affect the projectile motion of a car driving off a cliff are the initial velocity, angle of launch, and the presence of air resistance. The initial velocity is the speed at which the car is moving when it leaves the edge of the cliff, and the angle of launch determines the direction in which the car will travel. Air resistance can also play a role in slowing down the car's motion.

2. How does the height of the cliff impact the projectile motion of the car?

The height of the cliff will affect the duration of the car's flight and the distance it travels. The higher the cliff, the longer the car will be in the air and the farther it will travel horizontally. This is because the car will have more time to accelerate due to the force of gravity.

3. Can the shape and weight of the car affect its projectile motion?

Yes, the shape and weight of the car can affect its projectile motion. A heavier car will have more inertia and will be more difficult to accelerate, resulting in a shorter horizontal distance traveled. The shape of the car can also impact air resistance, with more aerodynamic cars experiencing less resistance and traveling farther.

4. What is the difference between horizontal and vertical velocity in projectile motion?

Horizontal velocity refers to the speed at which the car is moving in the horizontal direction, while vertical velocity refers to the speed at which the car is moving in the vertical direction. In projectile motion, the horizontal velocity remains constant, while the vertical velocity changes due to the force of gravity.

5. How does the angle of launch impact the trajectory of the car in projectile motion?

The angle of launch determines the direction in which the car will travel. A higher angle will result in a steeper trajectory, while a lower angle will result in a flatter trajectory. The optimal angle for maximum distance in projectile motion is 45 degrees.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
798
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
804
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
20K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top