# Projectile motion of car driving off a cliff

1. Aug 6, 2013

### Alyssa Jesse

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 :)

2. Aug 6, 2013

### voko

This is correct.

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

What can you say about the vertical velocity when the maximum height is reached?

3. Aug 6, 2013

### Alyssa Jesse

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. Aug 6, 2013

### voko

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

Right. Now, does that allow you to compute when this will happen? And knowing when, can you compute the vertical distance?

5. Aug 6, 2013

### Alyssa Jesse

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. Aug 6, 2013

### Alyssa Jesse

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

7. Aug 6, 2013

### voko

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

No, this cannot be correct. Dimensionally this is velocity, while you need distance.

8. Aug 6, 2013

### Alyssa Jesse

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. Aug 6, 2013

### voko

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. Aug 6, 2013

### Alyssa Jesse

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. Aug 6, 2013

### voko

How can (1/2) * (15) * (1.5) be equal to 0.05?

12. Aug 6, 2013

### Alyssa Jesse

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. Aug 6, 2013

### voko

Looking good now!

14. Aug 7, 2013

### Alyssa Jesse

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. Aug 7, 2013

### voko

I would guess - guess - that the initial and final points of the jump are at an equal height.

16. Aug 7, 2013

### Alyssa Jesse

The problem is I'm not told what the initial and final heights of the jump are, so how can I express this?

17. Aug 7, 2013

### voko

As I said: I think they should be considered equal. In that case, does it really matter what their heights are?

18. Aug 7, 2013

### Alyssa Jesse

Hmm I understand the concept behind what you are saying, but I don't know how to show that using physics.

19. Aug 7, 2013

### voko

What does "making the jump" mean?

20. Aug 7, 2013

### Alyssa Jesse

That the car will reach the other side without falling into the abyss?