# Time and Displacement

1. Aug 16, 2007

### Iniuria12

Ok, I have an example problem that doesn't show any of the actual calculations and I'm stumped at the process of finding the answer. The Answers were given just not the work behind it.

Bill and Ted have a skateboarding contest. They begin side by side and push their boards forward. At a speed of 5 m/s they both jump straight up and then land on their moving skateboards. Bill goes 7.5m before landing and Ted goes 6.0m before landing.
a) How long was each boy in the air?
b) How high did each boy jump?

Vi=5m/s, D1=7.5m . D2=6.0m

so.. t=D/V... t=7.5m / 5m/s = 1.5s and t=6m / 5m/s = 1.2s
so Bill has airtime of 1.5s and Ted has airtime of 1.2s.
And they give they answer of Bill is 2.76m in the air and Ted is 1.76m in the air, but unsure of the calculations for these answers..

Any help is appreciated.
Thanks

2. Aug 16, 2007

### mgb_phys

Sorry - I misunderstood, does the question mean that they launch the scateboards at a speed of 5m/s or they jump at 5 m/s.

Otherwise they can't both jump at the same speed and same angle and have diffeent times in the air ( neglecting air resistance )

3. Aug 16, 2007

### Iniuria12

The question stated that at a speed of 5 m/s they both jump straight up and then land on their skateboards. Bills board goes 7.5 m before he lands and Teds goes 6.0 m before he lands. So I'm assuming it means the speed of the skateboard

4. Aug 16, 2007

### mgb_phys

Ok a little trickier but not much.
You know the time-of-flight of the two boys.

The time to reach the maximum height at the top of their trajectory is half this. (Neglecting air resistance)
You can use V = U + at to get their initial vertical speed.
Then you can use conservation of energy to get their maximum height.

ie, ke = 1/2 m v^2 = pe = mg h since the m cancels you don't need their mass and
h = (1/2 v^2 )/g where v is their initial vertical speed

5. Aug 16, 2007

### Iniuria12

That puts it into perspective.. Appreciate your help. Thanks

6. Aug 16, 2007

### learningphysics

Once you know the time to get to the top... there's a kinematic equation that will directly give you the height... you know the final velocity and time and acceleration...

7. Aug 16, 2007

### mgb_phys

I always like to use conservation of energy where you can in these sort of problems - it's not always enough to solve the whole question, but it is always right and avoids a lot of complications about forces and directions.

8. Aug 16, 2007

### Iniuria12

final velocity would be -9.7 m/s, time = 1.5s and 1.2s (two different boys), and the acceleration, I'm assuming since they jump directly upwards it would simply be g= -9.8m/s^2...

9. Aug 16, 2007

### learningphysics

Yeah there are multiple ways to solve the problem... maybe conservation of energy is the best way... you can also use:

s = v2*t - (1/2)at^2 (where s is the displacement while going up to the maximum height... where v2 is the velocity at the top which is 0... and using t is half the value calculated before)...

or you can use
s = v1*t + (1/2)at^2 (this is for the downward path coming back down from the top... where v1 is the velocity at the top which is 0...)

for both you'd use a=-g, and t is half of what you calculated before... the second equation gives the negative of the first, since it's downward displacement.

Last edited: Aug 16, 2007
10. Aug 16, 2007

### Iniuria12

:=).. Thank you both so much.. Finally it makes sense and comes together perfectly.. Appreciate the help both of you have given. Thanks

11. Aug 16, 2007

### learningphysics

You're welcome.

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