Hang Time for Skateboard Jumping

[killaI9BI]

Homework Statement

Two boys have a "hang time" contest on their skateboards. At a speed of 5.0 m/s, they both jump straight up and then land on their moving skateboards. The first boy goes a horizontal distance of 7.5 m before he lands, while the second boy goes 6.0 m before he lands.

a) How long was each boy in the air

b) How high did each boy jump? (Hint: Separate the horizontal and vertical motions. Time is the same for both)

Homework Equations

d = vi X t + 1/2a X t²

The Attempt at a Solution

a) first boy: 7.5/5 = 1.5 s
second boy: 6/5 = 1.2 s

b)
d = 5 X 1.5 + ½(-9.8) X 1.5²
d = 7.5 + (-3775)
d = 3.725 m

d = 5 X 1.2 + ½(-9.8) X 1.2²
d = 6 + (-7.056)
d = (-1.056) m

I don't know if I'm using the wrong formula or a wrong value... possibly because I calculated the answer to the first question incorrectly?

 

[gneill]

For (b), pay close attention to the given hint. Also, consider that the vertical motion can be broken into two parts: going up and coming back down. How is the hang time split between these two parts?

 

[killaI9BI]

I should've caught that. My answer should be 1/2 of the distances calculated above - I think.

If I did a) correctly and I can assume half of the hang time is rising and the other half is falling at a rate of (-9.8) m/s, then:

for the first boy:
d = ½(0 + (-9.8)) X 0.75
d = (-3.675)
d = 3.7 m

for the second boy:
d = ½(-9.8) X 0.6
d = (-2.94)
d = 2.9 m

Is that right?

 

[killaI9BI]

Re-done with the right formula (I think)

b)
first boy:

d = 0.75 + 1/2(-9.8) X 0.75²
d = (-2.0)m

second boy:

d = 0.6 + 1/2(-9.8) X 0.6²
d = (-1.16)
d = (-1.2) m

 

[gneill]

Re-done with the right formula (I think)

b)
first boy:

d = 0.75 + 1/2(-9.8) X 0.75²
d = (-2.0)m

What is the meaning of the first 0.75 value in your equation above? Perhaps you should write out the equation symbolically first before plugging in numbers; it will make following your logic easier.

Does it make sense that an upward jump should result in a negative height?

You have made the correct deduction that it is the time that needs to be halved: you only want to consider half of the trajectory. A simplifying choice is to consider the downward half, starting at the maximum height with zero vertical velocity. This way only the acceleration term is involved in producing a change in distance.

The choice of coordinate system is important. If an object is being dropped from some height and distance is considered to be increasing as it falls, then the y-axis is pointing downwards. Velocity, too, will increase as the object falls downwards. You should make your acceleration constant take the appropriate sign to match increasing distance and velocity.

 

[killaI9BI]

the first 0.75 is half of the air time that I calculated from a). Sorry, I should've made that clear so you can follow what I'm trying to do

I made the mistake of not changing my answers to positive values. I was calculating the height of the jump by considering the acceleration of gravity (-9.8) m/s² as well as half of the air time which I calculated in a) which is why the values were negative.

The answers should've been 2.0 m for the first boy and 1.2 m for the second boy. I shouldn't rush but I needed to finish my thought before having to leave for the day.

Thank you for your help. Would you say I have answered both questions correctly?

a) first boy: 7.5/5 = 1.5 s
second boy: 6/5 = 1.2 s

b) The first boy jumped 2.0 m high and the second boy jumped 1.2 m high.

 

[Nathanael]

d = 0.75 + 1/2(-9.8) X 0.75²

If 0.75 is the time, then you're implying that the initial velocity was 1m/s?

But the initial velocity is unknown.

 

[killaI9BI]

I don't have an excuse for my carelessness unfortunately

I sincerely apologize for wasting your time.

a) 7.5/5 = 1.5s
6/5 = 1.2s

b) d = vi X t + 1/2a X t²
d = 1/2(-9.8) X 0.75²
d = (-2.756)
d = 2.8 m

d = 1/2(-9.8) X 0.6²
d = (-1.764) m
d = 1.8 m

My interpretation of these equations would be calculating the height of the jump from the top to the time they reach the ground.

Initial velocity should be zero so the equations consider a = (-9.8) m/s² which is the acceleration of gravity and t = 1/2 of the air time calculated in a).

...I think.

 

[Nathanael]

I sincerely apologize for wasting your time.

What is time for but to be wasted

d = 2.8 m
...
d = 1.8 m

I agree with your answers.

My interpretation of these equations ...

The way I think of the equations (just to give you another perspective) is that $-gt$ is the change in velocity, but since the final velocity is zero, $gt$ must represent the initial velocity

Therefore $\frac{gt}{2}$ represents the average velocity, and so multiply that by time to get the height $d=V_{avg}t=\frac{gt}{2}t=4.9t^2$

But of course we must be careful to use $t=\frac{T}{2}$ where T is the total time in the air

[P.S. everything I said has to do with only the vertical component (of velocity and displacement)]

It is very good that you're thinking about the meaning of the equations. Too many are content with blindly using formulas.